case study questions on sequence and series class 11

CBSE Case Study Questions for Class 11 Maths Sequences and Series Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Sequences and Series  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 11 Maths Sequences and Series PDF

Checkout our case study questions for other chapters.

• Chapter 7 Permutations and Combinations Case Study Questions
• Chapter 8 Binomial Theorem Case Study Questions
• Chapter 10 Straight Lines Case Study Questions
• Chapter 11 Conic Sections Case Study Questions

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Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

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Class 11 Mathematics Case Study Questions

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If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

I.	Sets and Functions	60	23
II.	Algebra	50	25
III.	Coordinate Geometry	50	12
IV.	Calculus	40	08
V.	Statistics and Probability	40	12
	Internal Assessment

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

1.	Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers.	44	55
	Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas
2	Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.	20	25
3	Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations	16	20
	Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria.

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

Periodic Tests (Best 2 out of 3 tests conducted)	10 Marks
Mathematics Activities	10 Marks

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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Chapter 8 Class 11 Sequences and Series

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Updated for new NCERT - 2023-2024 Edition.

Solutions of Chapter 8 Sequences and Series of Class 11 NCERT book available free. All exercise questions, examples, miscellaneous are done step by step with detailed explanation for your understanding. 

In this Chapter we learn about Sequences

Sequence is any group of numbers with some pattern.

Like 2, 4, 8, 16, 32, 64, 128, 256, ....

1, 2, 3, 4, 5, 6, 7, 8

In this chapter we learn

• What a sequence is - and what is finite, infinite sequence, terms of a sequence
• What a series is - it is the sum of a sequence
• Denoting Sum by sigma Σ , and what it means
• What is an AP (Arithmetic Progression) is, and finding its n th term and sum
• Inserting AP between two numbers
• What is Arithmetic Mean ( AM ), and how to find it
• What is a GP (Geometric Progression) is, and finding its n th term and sum
• Inserting GP between two numbers
• What is Geometric Mean ( GM ), and how to find it
• Relationship between AM and GM
• Sum of special series
• Finding sum of series when n th term is given
• Finding sum of series when n th term is not given

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CBSE Class 11 Maths – Chapter 9 Sequences and Series- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sequence and Series : Notes and Study Materials -pdf

• Concepts of  Sequence and Series
• Sequence and Series Master File
• Sequence and Series Revision Notes
• R D Sharma Solution of AP
• R D Sharma Solution of GP
• R D Sharma Solution of Special Series
• NCERT Solution  Sequence and Series
• NCERT  Exemplar Solution Sequence and Series
• Sequence and Series : Solved Example 1

It means an arrangement of number in definite order according to some rule

Important point

• The various numbers occurring in a sequence are called its terms. The number of terms is called the length of the series
• Terms of a sequence are denoted generally by $a_1 , a_2, a_3, ….., a_n$ etc., The subscripts denote the position of the term. We can choose any other letter to denote it
• The nth term is the number at the nth position of the sequence and is denoted by a n
• The nth term is also called the general term of the sequence
• A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it of the type {1, 2, 3…k}.
• Many times is possible to express general term in terms of algebraic formula. But it may not be true in other cases. But we should be able to generate the terms of the sequence using some rules or theoretical scheme

Finite sequence A sequence containing a finite number of terms is called Finite sequence infinite sequence A sequence is called infinite if it is not a finite sequence.

Let a 1 , a 2 , a 3 ..be the sequence, then the sum expressed a 1 + a 2 +a 3 + ……. is called series.

• A series is called finite series if it has got finite number of terms
• A series is called infinite series if it has got infinite terms
• Series are often represented in compact form, called sigma notation, using the Greek letter σ (sigma)
• so , a 1 + a 2 +a 3 + …….a n can be expressed as $\sum_{k=1}^{n}a_k$

Example $1+2+3 + 4+ 5+….+100$ $= \sum_{k=1}^{n}k$

Types of Sequences

Arithmetic progression.

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant Examples 1. $1,5,9,13,17….$ 2. $1,2,3,4,5,…..$ Important Notes about Arithmetic Progression 1.The difference between any successive members is a constant and it is called the common difference of AP 2. If a 1 , a 2 ,a 3 ,a 4 ,a 5 are the terms in AP then D=a 2 -a 1 =a 3 – a 2 =a 4 – a 3 =a 5 –a­ 4 3. We can represent the general form of AP in the form a,a+d,a+2d,a+3d,a+4d.. Where a is first term and d is the common difference 4. Nth term of Arithmetic Progression is given by n th term = a + (n – 1)d 5. Sum of nth item in Arithmetic Progression is given by $S_n =\frac {n}{2}[a + (n-1)d]$ Or $S_n =\frac {n}{2}[t_1+ t_n]$ Arithmetic Mean The arithmetic mean A of any two numbers a and b is given by (a+b)/2 i.e. a, A, b are in AP A -a = b- A or $A = \frac {(a+b)}{2}$ If we want to add n terms between A and B so that result sequence is in AP Let A 1 , A 2 , A 3 , A 4 , A 5 …. A n be the terms added between a and b Then b= a+ [(n+2) -1]d or $d = \frac {(b-a)}{(n+1)}$ So terms will be $a+ \frac {(b-a)}{(n+1)}, a + 2 \frac {(b-a)}{(n+1)} ,……. a+ n \frac {(b-a)}{(n+1)}$

Geometric Progressions

A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout. Examples 1. $2,4,8,16,32….$ 2. $3,6,12,24,48,…..$ Important Notes about Geometric Progressions

• The ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio.
• If a 1 , a 2 ,a 3 ,a 4 ,a 5 are the terms in AP then $r=\frac {a_2}{a_1} =\frac {a_3}{a_2} =\frac {a_4}{a_3}=\frac {a_5}{a_4}$
• We can represent the general form of G.P in the form a,ar,ar 2 ,ar 3 …… Where a is first term and r is the common ratio
• Nth term of Geometric Progression is given by n th term = ar n-1
• Sum of nth item inGeometric Progression is given by $S_n =a \frac {r^n -1}{r-1}$ Or $S_n =a \frac {1-r^n }{ 1-r}$

Geometric Means The Geometric mean G of any two numbers a and b is given by (ab) 1/2 i.e. a, G, b are in G.P $\frac {G}{a} = \frac {b}{G}$ or $G=\sqrt {ab}$ If we want to add n terms between A and B so that result sequence is in GP Let G 1 , G 2 , G 3 , G 4 , G 5 …. G n be the terms added between a and b Then $b= ar^{n+1}$ or $r= (\frac {b}{a})^{\frac {1}{n+1}}$ So terms will be $a(\frac {b}{a})^{\frac {1}{n+1}}, a(\frac {b}{a})^{\frac {2}{n+1}} ,……. a(\frac {b}{a})^{\frac {n}{n+1}}$ Relationship Between A.P and G.P Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively. Then A ≥ G And $A – G = \frac {{\sqrt {a} – \sqrt {b}}^2}{2}$

Sum of Special Series

Sum of first n natural numbers $1 + 2 + 3 +….. + n = \frac {n(n+1}{2}$ $ \sum_{k=1}^{n} n =\frac {n(n+1}{2}$ sum of squares of the first n natural numbers $1^2+ 2^2+3^2 +…. + n^2 = \frac {n(n+1)(2n+1)}{6}$ $ \sum_{k=1}^{n} n^2= \frac {n(n+1)(2n+1)}{6}$ Sum of cubes of the first n natural numbers $1^3+ 2^3+3^3 +….. + n^3 = \frac {(n(n+1))^2}{4}$ $ \sum_{k=1}^{n} n^3= \frac {(n(n+1))^2}{4}$

Rules for finding the sum of Series

a. Write the nth term T n of the series b. Write the T n in the polynomial form of n $T_n= a n^3 + bn^2 + cn +d$ c. The sum of series can be written as $ \sum S = a \sum {n^3} + b \sum {n^2} + c \sum n + nd We already know the values if these standard from the formula given above and we can easily find the sum of the series Example Find the sum of the series $2^2 + 4^2 + 6^2 + …..(2n)^2} Solution Let nth term T n of the series Then $T_n = (2n)^2 = 4n^2$ Now $2^2 + 4^2 + 6^2 + …..(2n)^2 = \sum_{k=1}^{n} 4k^2 = 4 sum_{k=1}^{n} k^2 = 4 \frac {n(n+1)(2n+1)}{6}$ $=\frac {2n(n+1)(2n+1)}{3}$

Method of Difference

Many times , nth term of the series can be determined. For example 5 + 11 + 19 + 29 + 41…… If the series is such that difference between successive terms are either in A.P or G.P, then we can the nth term using method of difference $S_n = 5 + 11 + 19 + 29 + … + a_{n-1} + a_n$ or $S_n= 5 + 11 + 19 + … + + a_{n-1} + a_n$ On subtraction, we get $0 = 5 + [6 + 8 + 10 + 12 + …(n – 1) terms] – a_n$ Here 6,8,10 is in A.P,So $a_n = 5 + \frac {n-1}{2} [12 + (n-1)2]$ or $ a_n= n^2 + 3n + 1$ Now it is easy to find the Sum of the series $S_n = \sum_{k=1}^{n} {k^2 +3k +1}$ $=\frac {n(n+2)(n+4)}{3}$

Sequences and Series Class 11 MCQs Questions with Answers

Question 1. If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in (a) AP (b) GP (c) HP (d) none of these

Answer: (a) AP Hint: Given a, b, c are in GP ⇒ b² = ac ⇒ b² – ac = 0 So, ax² + 2bx + c = 0 have equal roots. Now D = 4b² – 4ac and the root is -2b/2a = -b/a So -b/a is the common root. Now, dx² + 2ex + f = 0 ⇒ d(-b/a)² + 2e×(-b/a) + f = 0 ⇒ db2 /a² – 2be/a + f = 0 ⇒ d×ac /a² – 2be/a + f = 0 ⇒ dc/a – 2be/a + f = 0 ⇒ d/a – 2be/ac + f/c = 0 ⇒ d/a + f/c = 2be/ac ⇒ d/a + f/c = 2be/b² ⇒ d/a + f/c = 2e/b ⇒ d/a, e/b, f/c are in AP

Question 2. If a, b, c are in AP then (a) b = a + c (b) 2b = a + c (c) b² = a + c (d) 2b² = a + c

Answer: (b) 2b = a + c Hint: Given, a, b, c are in AP ⇒ b – a = c – b ⇒ b + b = a + c ⇒ 2b = a + c

Question 3: Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is (a) 2 + √3 (b) 2 – √3 (c) 2 ± √3 (d) None of these

Answer: (a) 2 + √3 Hint: Let the three numbers be a/r, a, ar Since the numbers form an increasing GP, So r > 1 Now, it is given that a/r, 2a, ar are in AP ⇒ 4a = a/r + ar ⇒ r² – 4r + 1 = 0 ⇒ r = 2 ± √3 ⇒ r = 2 + √3 {Since r > 1}

Question 4: The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is (a) n/(n+1) (b) 1/(n+1) (c) 1/n (d) None of these

Answer: (a) n/(n+1) Hint: Given series is: S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1) ⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1)) ⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1) ⇒ S = 1 – 1/(n+1) ⇒ S = (n + 1 – 1)/(n+1) ⇒ S = n/(n+1)

Question 5: If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then (a) a, b, c are in AP (b) a², b², c² are in AP (c) 1/1, 1/b, 1/c are in AP (d) None of these

Answer: (b) a², b², c² are in AP Hint: Given, 1/(b + c), 1/(c + a), 1/(a + b) ⇒ 2/(c + a) = 1/(b + c) + 1/(a + b) ⇒ 2b² = a² + c² ⇒ a², b², c² are in AP

Question 6: The sum of series 1/2! + 1/4! + 1/6! + ….. is (a) e² – 1 / 2 (b) (e – 1)² /2 e (c) e² – 1 / 2 e (d) e² – 2 / e

Answer: (b) (e – 1)² /2 e Hint: We know that, e x = 1 + x/1! + x² /2! + x³ /3! + x 4 /4! + ……….. Now, e 1 = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ……….. e -1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ……….. e 1 + e -1 = 2(1 + 1/2! + 1/4! + ………..) ⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..) ⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..) ⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ……….. ⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ……….. ⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ……….. ⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Question 7: The third term of a geometric progression is 4. The product of the first five terms is (a) 4 3 (b) 4 5 (c) 4 4 (d) none of these

Answer: (b) 4 5 Hint: here it is given that T 3 = 4. ⇒ ar² = 4 Now product of first five terms = a.ar.ar².ar³.ar 4 = a 5 r 10 = (ar 2 ) 5 = 4 5

Question 8: Let Tr be the r th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n equals (a) 1/m n (b) 1/m + 1/n (c) 1 (d) 0

Answer: (c) 1 Hint: Let first term is a and the common difference is d of the AP Now, T m = 1/n ⇒ a + (m-1)d = 1/n ………… 1 and T n = 1/m ⇒ a + (n-1)d = 1/m ………. 2 From equation 2 – 1, we get (m-1)d – (n-1)d = 1/n – 1/m ⇒ (m-n)d = (m-n)/mn ⇒ d = 1/mn From equation 1, we get a + (m-1)/mn = 1/n ⇒ a = 1/n – (m-1)/mn ⇒ a = {m – (m-1)}/mn ⇒ a = {m – m + 1)}/mn ⇒ a = 1/mn Now, T mn = 1/mn + (mn-1)/mn ⇒ T mn = 1/mn + 1 – 1/mn ⇒ T mn = 1

Question 9. The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is (a) 2 (b) 4 (c) 6 (d) 8

Answer: (c) 6 Hint: Let a and b are two numbers such that a + b = 13/6 Let A 1 , A 2 , A 3 , ………A 2n be 2n arithmetic means between a and b Then, A 1 + A 2 + A 3 + ………+ A 2n = 2n{(n + 1)/2} ⇒ n(a + b) = 13n/6 Given that A 1 + A 2 + A 3 + ………+ A 2n = 2n + 1 ⇒ 13n/6 = 2n + 1 ⇒ n = 6

Question 10. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in (a) A.P. (b) G.P. (c) H.P. (d) A.G.P.

Answer: (c) H.P. Hint: Given, equation is ax² + bx + c = 0 Let p and q are the roots of this equation. Now p+q = -b/a and pq = c/a Given that p + q = 1/p² + 1/q² ⇒ p + q = (p² + q²)/(p² ×q²) ⇒ p + q = {(p + q)² – 2pq}/(pq)² ⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)² ⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a} ⇒ -bc²/a³ = {b² – 2ca}/a² ⇒ -bc²/a = b² – 2ca Divide by bc on both side, we get ⇒ -c /a = b/c – 2a/b ⇒ 2a/b = b/c + c/a ⇒ b/c, a/b, c/a are in AP ⇒ c/a, a/b, b/c are in AP ⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP ⇒ a/c, b/a, c/b are in HP

Question 11. If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then (a) a, b, c are in AP (b) a², b², c² are in AP (c) 1/1, 1/b, 1/c are in AP (d) None of these

Question 12. The 35th partial sum of the arithmetic sequence with terms a n = n/2 + 1 (a) 240 (b) 280 (c) 330 (d) 350

Answer: (d) 350 Hint: The 35th partial sum of this sequence is the sum of the first thirty-five terms. The first few terms of the sequence are: a 1 = 1/2 + 1 = 3/2 a 2 = 2/2 + 1 = 2 a 3 = 3/2 + 1 = 5/2 Here common difference d = 2 – 3/2 = 1/2 Now, a 35 = a 1 + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2 Now, the sum = (35/2) × (3/2 + 37/2) = (35/2) × (40/2) = (35/2) × 20 = 35 × 10 = 350

Question 13. The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is (a) 2 (b) 4 (c) 6 (d) 8

Question 14. The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (c) 3 Hint: Let first term of the GP is a and common ratio is r. 3rd term = ar² 5th term = ar 4 Now ⇒ ar² + ar 4 = 90 ⇒ a(r² + r 4 ) = 90 ⇒ r² + r 4 = 90 ⇒ r² ×(r² + 1) = 90 ⇒ r²(r² + 1) = 3² ×(3² + 1) ⇒ r = 3 So the common ratio is 3

Question 15. The sum of AP 2, 5, 8, …..up to 50 terms is (a) 3557 (b) 3775 (c) 3757 (d) 3575

Answer: (b) 3775 Hint: Given, AP is 2, 5, 8, …..up to 50 Now, first term a = 2 common difference d = 5 – 2 = 3 Number of terms = 50 Now, Sum = (n/2)×{2a + (n – 1)d} = (50/2)×{2×2 + (50 – 1)3} = 25×{4 + 49×3} = 25×(4 + 147) = 25 × 151 = 3775

Question 16. If 2/3, k, 5/8 are in AP then the value of k is (a) 31/24 (b) 31/48 (c) 24/31 (d) 48/31

Answer: (b) 31/48 Hint: Given, 2/3, k, 5/8 are in AP ⇒ 2k = 2/3 + 5/8 ⇒ 2k = 31/24 ⇒ k = 31/48 So, the value of k is 31/48

Question 17. The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is (a) n/(n+1) (b) 1/(n+1) (c) 1/n (d) None of these

Answer: (a) n/(n+1) Hint: Given series is: S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1) ⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1)) ⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1) ⇒ S = 1 – 1/(n+1) ⇒ S = (n + 1 – 1)/(n+1) ⇒ S = n/(n+1)

Question 18. If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is (a) 228 (b) 74 (c) 740 (d) 1090

Answer: (c) 740 Hint: Let a is the first term and d is the common difference of AP Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term ⇒ a + 2d = 7 ………….. 1 and 3(a + 2d) + 2 = a + 6d ⇒ 3×7 + 2 = a + 6d ⇒ 21 + 2 = a + 6d ⇒ a + 6d = 23 ………….. 2 From equation 1 – 2, we get 4d = 16 ⇒ d = 16/4 ⇒ d = 4 From equation 1, we get a + 2×4 = 7 ⇒ a + 8 = 7 ⇒ a = -1 Now, the sum of its first 20 terms = (20/2)×{2×(-1) + (20-1)×4} = 10×{-2 + 19×4)} = 10×{-2 + 76)} = 10 × 74 = 740

Question 19. If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals (a) 10 (b) 12 (c) 11 (d) 13

Answer: (c) 11 Hint: Given, the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the A.P. 57, 59, 61, …. ⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2} ⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2} ⇒ 6n + 1 = {2n + 112}/2 ⇒ 6n + 1 = n + 56 ⇒ 6n – n = 56 – 1 ⇒ 5n = 55 ⇒ n = 55/5 ⇒ n = 11

Question 20. If a is the A.M. of b and c and G 1 and G 2 are two GM between them then the sum of their cubes is (a) abc (b) 2abc (c) 3abc (d) 4abc

Answer: (b) 2abc Hint: Given, a is the A.M. of b and c ⇒ a = (b + c) ⇒ 2a = b + c ………… 1 Again, given G 1 and G 1 are two GM between b and c, ⇒ b, G 1 , G 2 , c are in the GP having common ration r, then ⇒ r = (c/b) 1/(2+1) = (c/b) 1/3 Now, G 1 = br = b×(c/b) 1/3 and G 1 = br = b×(c/b) 2/3 Now, (G 1 )³ + (G 2 )3 = b³ ×(c/b) + b³ ×(c/b)² ⇒ (G 1 )³ + (G 2 )³ = b³ ×(c/b)×( 1 + c/b) ⇒ (G 1 )³ + (G 2 )³ = b³ ×(c/b)×( b + c)/b ⇒ (G 1 )³ + (G 2 )³ = b² ×c×( b + c)/b ⇒ (G 1 )³ + (G 2 )³ = b² ×c×( b + c)/b ………….. 2 From equation 1 2a = b + c ⇒ 2a/b = (b + c)/b Put value of(b + c)/b in eqaution 2, we get (G 1 )³ + (G 2 )³ = b² × c × (2a/b) ⇒ (G 1 )³ + (G 2 )³ = b × c × 2a ⇒ (G 1 )³ + (G 2 )³ = 2abc

(Education-Your Door To The Future)

CBSE Class 11 Term-1 Maths 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

CBSE Class 11 Term-1 Maths 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

In this Post I Have Provided Mathematics  Class 11 Chapter-Wise MCQs, Case-Based Question And Assertion/Reason For Term-1 Session 2021-22 With Solutions. CBSE has Recently Included These Types Of MCQs  And Assertion/Reason For Term-1 Exam 2020.   Every Student Knows these types of Questions Are Very Important For Their Term-1 Examination.

Any student who want to download these MCQs, then they have to click on given respective download links and it will be automatically download in your Google Drive.

CBSE Class 11th  Physics  Term-1 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

For Downlod   Click Here

Given Below Are The Class 11 Mathematics Chapter Name with Respective Download Links Containing Study Material:

1. Sets
2. Relations and Functions
3. Complex Numbers
4. Sequence and Series
5. Straight Lines
6. Limits
7. Statistics

CBSE Sample Papers For Class 11 Mathematics

CBSE sample paper for class  11th Maths  is an important tools to analyse itself. Before going to final examinations every student try to solve different types of sample question paper. It will enhance your knowledge and also provide to increase mental ability. Sample question paper are the best resources for the student to prepare for their Board examinations.

These sample question papers are very helpful to the student to get an entire experience before attempting the Board examinations papers. If students solve different types of sample question papers then they get the good confidence about the appropriate answer and they make good score in their Board examinations.

As every student know CBSE always change their exam pattern and also the model paper for their Board examinations that is why it is very important to all the student to understand the Model paper pattern before going to the final examinations.

In this articles I have provided different types of question papers on the basis of CBSE latest exam pattern which will definitely help to the student for the good preparation for their examinations.

In every year before the Board examination, CBSE provide model test paper to the student, on the basis of these model test papers I have prepared sample question papers for class  11th Maths  which will help to the student to understand the concept of model test paper and also it will help to the student for good preparation for their board examination.

CBSE Sample Paper:

Maths And Physics With Pandey Sir  is a website which provide free study material and notes to every students who are preparing for their Board examinations. By going in Level Section you can select CBSE sample papers. In this section, I have provided CBSE sample paper for class 9th to 12th.

Also if you are preparing for any competitive exams, they can find out study material and important notes in Level Section. In this website I have also provided most important MCQs questions, Assertion/Reason based questions and case study based questions for Board examinations or any other competitive examinations.

Benefits Of Solving CBSE Sample Papers for Examination:

Time Management:-

With the help of CBSE sample question papers you can make a proper strategy for their examinations also by practicing sample question papers you can manage your time during the Board examinations.

Examination Strategy:-

After practicing different types of sample question papers you can understand on which section you have spend more time and which section is easy and which section is hard. In this way you can make a proper examination strategy.

Self Evaluation:-

Self evaluation is very important to every student because with the help of self evaluation, student can understand their weak and strong point, which is very important for any Board examinations. After practicing more and more simple paper you can evaluate himself.

To Identify Silly Mistakes:-

As every student know there are many questions in their question paper which are very easy  but student make mistakes in this types of question. By solving different types of sample question papers you can reduce their silly mistakes.

Advantages Of CBSE Sample Question Papers:

* To get a good exam ideas about your examination paper patterns

* Student can easily access to the question papers

* It is very helpful to know about the mark distribution system.

* Helpful in self evaluation

* To know about the pattern of CBSE

* Help to adjust a proper time management

* Available in PDF format

* Chapter wise CBSE sample papers

* CBSE sample papers for class 10

* CBSE sample paper class 12

* CBSE sample paper 2021-22 class 12 Solutions

* CBSE sample paper class 11

* CBSE Sample paper class 9

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

Ex 9.1 Class 11 Maths 

Sequences and Series Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series All Exercises were prepared by Experienced LearnCBSE.in Teachers.

Free download NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1, Ex 9.2, Ex 9.3, Ex 9.4 and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

• Sequences and Series Class 11 Ex 9.1
• Sequences and Series Class 11 Ex 9.2
• Sequences and Series Class 11 Ex 9.3
• Sequences and Series Class 11 Ex 9.4
• Sequences and Series Class 11 Miscellaneous Exercise
• अनुक्रम तथा श्रेणी प्रश्नावली 9.1 का हल हिंदी में
• अनुक्रम तथा श्रेणी प्रश्नावली 9.2 का हल हिंदी में
• अनुक्रम तथा श्रेणी प्रश्नावली 9.3 का हल हिंदी में
• अनुक्रम तथा श्रेणी प्रश्नावली 9.4 का हल हिंदी में
• अनुक्रम तथा श्रेणी विविध प्रश्नावली का हल हिंदी में
• Sequences and Series Class 11 Notes
• NCERT Exemplar Class 11 Maths Sequences and Series
• JEE Main Mathematics Sequences and Series Previous Year Questions

Topics and Sub Topics in Class 11 Maths Chapter 9 Sequences and Series:

9	Sequences and Series
9.1	Introduction
9.2	Sequences
9.3	Series
9.4	Arithmetic Progression (A.P.)
9.5	Geometric Progression (G.P.)
9.6	Relationship Between A.M. and G.M.
9.7	Sum to n terms of Special Series

 NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1

Ex 9.1 Class 11 Maths Question 1: Write the first five terms of the sequence whose nth term is a n  = n (n + 2). Ans: a n  = n(n +2) Substituting n = 1, 2, 3, 4 and 5, we obtain a 1  = 1 (1 + 2) = 3, a 2  = 2 (2 + 2) = 8, a 3  = 3 (3 + 2) = 15, a 4  = 4 (4 + 2) = 24, a 5  = 5 (5 + 2) = 35 Therefore, the required terms are 3, 8, 15, 24 and 35.

Ex 9.1 Class 11 Maths Question 2: Write the first five terms of the sequence whose nth term is a n = \(\frac{n}{n+1}\). Ans:

a n = \(\frac{n}{n+1}\) Sustituting n = 1, 2, 3, 4, 5, we otain a n = \(\frac{1}{1+1}=\frac{1}{2}\)

a n = \(\frac{2}{2+1}=\frac{2}{3}\)

a n = \(\frac{3}{3+1}=\frac{3}{4}\)

a n = \(\frac{4}{4+1}=\frac{4}{5}\)

a n = \(\frac{5}{5+1}=\frac{5}{6}\)

Therefore, the required terms are \(\frac{1}{2}\), \(\frac{2}{3}\) , \(\frac{3}{4}\) , \(\frac{4}{5}\) and \(\frac{5}{6}\) .

More Resources for CBSE Class 11

• RD Sharma Class 11 Solutions Some Special Series

NCERT Solutions

• NCERT Solutions Class 11 Maths
• NCERT Solutions Class 11 Physics
• NCERT Solutions Class 11 Chemistry
• NCERT Solutions Class 11 Biology
• NCERT Solutions Class 11 Hindi
• NCERT Solutions Class 11 English
• NCERT Solutions Class 11 Business Studies
• NCERT Solutions Class 11 Computer Science

Ex 9.1 Class 11 Maths Question 3: Write the first five terms of the sequence whose nth term is a n = 2 n . Ans: a n = 2 n Substituting n = 1, 2, 3, 4, 5, we obtain a 1 = 2 1 = 2 a 2 = 2 2 = 4 a 3 = 2 3 = 8 a 4 = 2 4 = 16 a 5 = 2 5 = 32 Therefore, the required terms are 2, 4, 8, 16 and 32.

Ex 9.1 Class 11 Maths Question 11: Write the first five terms of the following sequence and obtain the corresponding series: a 1 = 3, a n = 3a n – 1 + 2 for all n > 1. Ans:

a 1 = 3, a n = 3 a n – 1 + 2 for all n > 1 => a 2 = 3a 2 – 1 + 2 = 3a 1 + 2 = 3(3) + 2 = 11

a 3 = 3a 3 – 1 + 2 = 3a 2 + 2 = 3(11) + 2 = 35

a 4 = 3a 4 – 1 + 2 = 3a 3 + 2 = 3(35) + 2 = 107

a 5 = 3a 5 – 1 + 2 = 3a 4 + 2 = 3(107) + 2 = 323. Hence, the first five terms of the sequence are 3, 11, 35, 107 and 323. The corresponding series is 3 + 11 + 35 + 107 + 323 + …

Ex 9.1 Class 11 Maths Question 13: Write the first five terms of the following sequence and obtain the corresponding series: a 1 = a 2 = 2, a n = a n – 1 – 1, n > 2. Ans:

a 1 = a 2 = 2, a n = a n – 1 , n > 2 => a 3 = a 2 – 1 = 2 – 1 = 1, a 4 = a 3 – 1 = 1 – 1 = 0 a 5 = a 4 – 1 = 0 – 1 = – 1. Hence, the first five terms of the sequence are 2, 2, 1, 0 and – 1. The corresponding series is 2 + 2 +1 + 0+ (- 1) + ……….

Ex 9.1 Class 11 Maths Question 14: The Fibonacci sequence is defined by 1 = a 1 = a 2 and a n = a n – 1 + a n – 2 n > 2. Find \(\frac{\boldsymbol{a}_{n+1}}{\boldsymbol{a}_{n}}\), for n = 1, 2, 3, 4, 5. Ans:

1 = a 1 = a 2 a n = a n – 1 + a n – 2 , n > 2 a 3 = a 2 + a 1 = 1 + 1 = 2, a 4 = a 3 + a 2 = 2 + 1 = 3, a 5 = a 4 + a 3 = 3 + 2 = 5, a 6 = a 5 + a 3 = 5 + 3 = 8

For n = 1, \(\frac{a_{n+1}}{a_{n}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}\) = 1

For n = 2, \(\frac{a_{n+1}}{a_{n}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}\) = 2

For n = 3, \(\frac{a_{n+1}}{a_{n}}=\frac{a_{4}}{a_{3}}=\frac{3}{2}\)

For n = 4, \(\frac{a_{n+1}}{a_{n}}=\frac{a_{5}}{a_{4}}=\frac{5}{3}\)

For n = 5, \(\frac{a_{n+1}}{a_{n}}=\frac{a_{6}}{a_{5}}=\frac{8}{5}\)

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (अनुक्रम तथा श्रेणी) Hindi Medium Ex 9.1

NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2

Ex 9.2 Class 11 Maths Question 1: Find the sum of odd integers from 1 to 2001. Ans: The odd integers from 1 to 2001 are 1, 3, 5, … 1999, 2001. This sequence forms an A.P. Here, first term, a = 1 Common difference, d = 2 Here, a + (n – 1) d = 2001 ⇒ 1 + (n -1) (2) = 2001 ⇒ 2n – 2 = 2000 ⇒ n = 1001 S n = \(\frac{n}{2}\) [2a + (n – 1)d] = \(\frac{1001}{2}\) [2 × 1 + (1001 – 1) × 2] = \(\frac{1001}{2}\) [2 + 1000 x 2] = \(\frac{1001}{2}\) × 2002 = 1001 × 1001 = 1002001 Thus, the sum of odd numbers from 1 to 2001 is 1002001.

Ex 9.2 Class 11 Maths Question 3: In an A.P. the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is – 112. Ans: Let the first term of given AP be a and common difference be d. We have, T 1 = a = 2 T 1 + T 2 + T 3 + T 4 + T 5 = [T 6 + T 7 + T 8 + T 9 + T 10 ] Sum of 5 terms, where first term is a = \(\frac{1}{4}\) × sum of 5 terms, where first term is (a + 5d) ⇒ \(\frac{5}{2}\) [2a + (5 – 1) d] = \(\frac{1}{4}\) × \(\frac{5}{2}\) [2(a + 5d) + (5 – 1)d] [∵ S n = \(\frac{n}{2}\) [2a + (n – 1)d]] \(\frac{5}{2}\) [2a + 4d] = \(\frac{1}{4}\) × \(\frac{5}{2}\) [2a + 10d + 4d] \(\frac{5}{2}\) [2a + 4d] = \(\frac{1}{4}\) × \(\frac{5}{2}\) [2a + 14d] ⇒ 2a + 4d = \(\frac{1}{4}\) [2a + 14d] 2(2) + 4d = \(\frac{1}{4}\) [2 . (2) + 14d] [put a = 2] 4 + 4d = \(\frac{1}{4}\) [4 + 14d] 16 + 16d = 4 + 14d 16d – 14d = 4 – 16 2d = – 12 d = – 6 T 20 = a + (20 – 1) d = 2 + 19 × (- 6) = 2 – 114 = – 112 Hence proved.

Ex 9.2 Class 11 Maths Question 6: If the sum of a certain number, of terms of the A.P. 25, 22, 19, …is 116. Find the last term. Ans: Let the sum of n terms of the given A.P. be 116. S n = \(\frac{n}{2}\) [2a + (n – 1)d] Here, a = 25 and d = 22 – 25 = – 3 S n = \(\frac{n}{2}\) [2 × 25 + (n – 1) (- 3)] ⇒ 116 = \(\frac{n}{2}\) [50 – 3n + 3] ⇒ 232 = n(53 – 3n) = 53n – 3n 2 3n 2 – 24n – 29n + 232 = 0 3n (n – 8) – 29 (n – 8) = 0 (n – 8) (3n – 29) = 0 n = 8 or n = \(\frac{29}{3}\) However, n cannot be equal to \(\frac{29}{3}\). Therefore, n = 8 ∴ a 8 = Last term = a + (n -1) d = 25 + (8 – 1) (- 3) = 25 + (7) (- 3) = 25 – 21 = 4 Thus, the last term of the A.P. is 4.

Ex 9.2 Class 11 Maths Question 10: If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find sum of the first (p + q) terms. Ans: Let a and b be the first term and the common difference of the A.P. respectively. Here, S p = \(\frac{p}{2}\) [2a + (p – 1)d] S q = \(\frac{q}{2}\) [2a + (q – 1 )d] According to the given condition, \(\frac{p}{2}\) [2a + (p – 1) d] = \(\frac{q}{2}\) [2a + (q – 1) d] p [2a + (p – 1) d] = q [2a + (n – 1) d] 2ap + pd (p – 1) d = 2aq + qd (q – 1) d 2a (p – q) + d[p(p – 1) – q (q – 1)] = 0 2a (p – q) + d[p 2 – p – q 2 + q] = 0 2a (p – q) + d [(p – q) (p + q) – (p – q)] = 0 2a (p – q) + d [(p – q) (p + q – 1)] = 0 2a + d (p + q – 1) = 0 d = \(\frac{-2 a}{p+q-1}\) ……………..(i) ∴ S p + q = \(\frac{p+q}{2}\) [2a + (p + q – 1) d] S p + q = \(\frac{p+q}{2}\left[2 a+(p+q-1)\left(\frac{-2 a}{p+q-1}\right)\right]\) [from eq. (i)] = \(\frac{p+q}{2}\) [2a – 2a] = 0 Thus, the sum of the first (p + q) terms of the A.P. is 0.

\(\frac{n^{2} d}{2}+n\left(a-\frac{d}{2}\right)\) = 3n 2 + 5n

Comparing the coefficient of n 2 on both sides, we obtain \(\frac{d}{2}\) = 3 ⇒ d 2 = 2 × 3 = 6 Comparing the coefficient of n on both sides, we obtain a – \(\frac{d}{2}\) = 5 a – \(\frac{6}{2}\) = 5 a = 5 + 3 = 8 Therefore, from eq. (i), we obtain 8 + (m – 1) 6 = 164 ⇒ (m – 1) 6 = 164 – 8 = 156 ⇒ m – 1 = 26 ⇒ m = 27 Thus the value of m is 27.

Ex 9.2 Class 11 Maths Question 18: The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon. Ans: The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°. It is known that the sum of all angles of a polygon with n sides is 180° (n – 2). ∵ S n = 180° (n – 2) \(\frac{n}{2}\) [2a + (n -1) d] = 180° (n – 2) \(\frac{n}{2}\) [240° + (n – 1) 5°] = 180 (n – 2) n [240 + (n – 1) 5] = 360 (n – 2) 240n + 5n 2 – 5n = 360n – 720 ⇒ 5n 2 + 235n – 360n + 720 = 0 ⇒ 5n 2 – 125n + 720 = 0 ⇒ n 2 – 25n + 144 = 0 ⇒ n 2 – 16n – 9n + 144 = 0 ⇒ n (n – 16) – 9 (n – 16) = 0 ⇒ (n – 9) (n – 16) = 0 ⇒ n = 9 or 16.

NCERT Solutions for class 11 Maths Chapter 9 Exercise 9.3

a 20 = ar 20 – 1

= \(\frac{5}{2}\left(\frac{1}{2}\right)^{19}\)

= \(\frac{5}{(2)(2)^{19}}=\frac{5}{(2)^{20}}\)

a n = ar n – 1

= \(\frac{5}{2}\left(\frac{1}{2}\right)^{n-1}\)

= \(\frac{5}{(2)(2)^{n-1}}=\frac{5}{(2)^{n}}\)

Ex 9.3 Class 11 Maths Question 10: Find the sum to n terms in the geometric progression x 3 , x 5 , x 7 … (if x ≠ ± 1). Ans: The given G.P. is x 3 , x 5 , x 7 , …………. Here, a = x 3 and r = x 2 S n = \(\frac{a\left(1-r^{n}\right)}{1-r}\)

= \(\frac{x^{3}\left[1-\left(x^{2}\right)^{n}\right]}{1-x^{2}}=\frac{x^{3}\left(1-x^{2 n}\right)}{1-x^{2}}\).

Ex 9.3 Class 11 Maths Question 12: The sum of first three terms of a G.P. is – and their product is 1. Find the common ratio and the terms. Ans: Let the first three number of G.P. be \(\frac{a}{r}\), a and ar. According to the question, \(\frac{a}{r}\) + a + ar = \(\frac{39}{10}\) … (i) and (\(\frac{a}{r}\)) × (a) × (ar) = 1 ⇒ a 3 = 1 ⇒ a = 1 On putting the value of a = 1 in eq. (i), we get \(\frac{1}{r}\) + 1 + r = \(\frac{39}{10}\) \(\frac{1+r+r^{2}}{r}=\frac{39}{10}\) ⇒ 10 + 10r + 10r 2 = 39r ⇒ 10r 2 + 10r – 39r + 10 = 0 ⇒ 10r 2 – 29r + 10 = 0 Now, factorising it by splitting the middle term, we get 10r 2 – 25r – 4r + 10 = 0 ⇒ 5r (2r – 5) – 2 (2r – 5) = 0 ⇒ (5r – 2)(2r – 5) = 0 ⇒ 5r – 2 = 0 and 2r – 5 = 0 ⇒ r = \(\frac{2}{5}\) and r = \(\frac{5}{2}\) When a = 1 and r = \(\frac{2}{5}\), then numbers are \(\frac{a}{r}=\frac{1}{\frac{2}{5}}=\frac{5}{2}\), a = 1 and ar = 1 × \(\frac{2}{5}\) = \(\frac{2}{5}\) ∴ \(\frac{5}{2}\), 1, \(\frac{2}{5}\).

When a = 1 and r = \(\frac{5}{2}\), then numbers are \(\frac{a}{r}=\frac{1}{5}=\frac{2}{5}\);

a = 1 and ar = 1 × \(\frac{5}{2}\) = \(\frac{5}{2}\) ∴ \(\frac{2}{5}\), 1, \(\frac{5}{2}\).

Ex 9.3 Class 11 Maths Question 13: How many terms of G.P. 3, 3 2 , 3 3 , …………… are needed to give the sum 120? Ans: The given G.P. is 3, 3 2 , 3 3 , ………… Let n terms of this G.P. be required to obtain the sum as 120. S n = \(\frac{a\left(r^{n}-1\right)}{r-1}\) Here, a = 3 and r = 3 S n = 120 = \(\frac{3\left(3^{n}-1\right)}{3-1}\)

⇒ 120 = \(\frac{3\left(3^{n}-1\right)}{3-1}\) ⇒ \(\frac{120 \times 2}{3}\) = 3 n – 1 ⇒ 3 n – 1 = 80 ⇒ 3 n = 81 ⇒ 3 n = 3 4 ∴ n = 4 Thus, four terms of the given G.P. are required to obtain the sum as 120.

Ex 9.3 Class 11 Maths Question 16: Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term. Ans: Let, first term = a and common ratio = r. Given, sum of first two terms = – 4 ⇒ a 1 + a 2 = – 4 ⇒ a + ar = – 4 ⇒ a (1 + r) = – 4 and fifth term = 4 × third term a 5 = 4 × a 3 ⇒ ar 4 = 4ar 2 r 2 = 4 r = ± 2 When r = 2, then from eq. (i), we get a (1 + 2) = – 4 ⇒ a = – \(\frac{4}{3}\) Then, G.P. is – \(\frac{4}{3}\), – \(\frac{4}{3}\) × 2, – \(\frac{4}{3}\) × (2) 2 , …………… i.e., \(\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}\) When r = – 2, then from eq. (i), we get a a (1 – 2)= – 4 ⇒ – a = – 4 ⇒ a = 4 Then, G.P. is 4, 4 × (- 2), 4 × (- 2) 2 … i.e, 4,- 8, 16, ………..

Ex 9.3 Class 11 Maths Question 17: If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P. Ans: Given, 4 th term, T 4 = x ⇒ ar 4 – 1 = x ⇒ ar 3 = x …………………..(i) 10th term, T 10 = y ⇒ ar 10 – 1 = y ⇒ ar 9 = y ………………….(ii) and 16 th term, T 16 = z ⇒ ar 16 – 1 = z ⇒ ar 15 = z ………………(iii) Now, multiplying eq. (j) by eq. (ill), we get ar 3 × ar 15 = x × z a 2 r 3 + 15 = x × z a 2 r 18 = xz (ar 9 ) 2 = xz ∴ y 2 = xz [from eq. (ii)] Therefore, x,y and z are in GP.

Ex 9.3 Class 11 Maths Question 20: Show that the products of the corresponding terms of the sequences a, ar, ar 2 , … ar n – 1 and A, AR, AR 2 , …….. AR n – 1 form a G.P. and find the common ratio. Ans: It has to be proved that the sequence, aA, arAR, ar 2 AR 2 , ………. ar n – 1 AR n – 1 , forms a G.P. \(\frac{\text { Second term }}{\text { First term }}=\frac{a r A R}{a A}\) = rR

\(\frac{\text { Third term }}{\text { Second term }}=\frac{a r^{2} A R^{2}}{a r A R}\) = rR Thus, the above sequence forms a G.P. and the common ratio is rR.

Ex 9.3 Class 11 Maths Question 21: Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4 th by 18. Ans: Let a be the first term and r the common ratio of G.P. ∴ nth term = T n = ar n – 1 ⇒ T 2 = ar , T 3 = ar 2 and T 4 = ar 3 Since third term is greater than the first by 9. ∴ T 3 = T 1 + 9 ⇒ ar 2 = a + 9 Second term is greater than the 4th by 18. T 2 = T 4 + 18 ⇒ ar = ar 3 + 18 ⇒ ar 3 = ar + 9r From eqs. (ii) and (iii), we get ar = ar + 9r + 18 ⇒ 0 = 9r + 18 ⇒ r = \(\frac{-18}{9}\) = – 2 Put = – 2 in (i), we get a(- 2) 2 = a + 9 ⇒ 4a = a + 9 ⇒ 3a = 9 ⇒ a = 3 T 2 = ar = 3 (- 2) = – 6 T 3 = ar 2 = 3 (- 2) 2 = 12 T 4 = ar 3 = 3 (- 2) 3 = – 24 ∴ Required terms are 3, – 6, 12 and – 24.

Thus, required ratio = \(\frac{a\left(1-r^{n}\right)}{(1-r)} \times \frac{(1-r)}{a r^{n}\left(1-r^{n}\right)}=\frac{1}{r^{n}}\)

Thus, the ratio of the sum of first n terms of G.P. to the sum of terms from (n + 1)th to (2n)th term is \(\frac{1}{r^{n}}\).

NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.4

Ex 9.4 Class 11 Maths Question 3: Find the sum of n terms of the series 3 × 1 2 + 5 × 2 2 + 7 × 3 2 + …………….. Ans:

The given series is 3 × 1 2 + 5 × 2 2 + 7 × 3 2 + …………….. nth term a n = (2n + 1) n 2 = 2n 3 + n 2 ∴ S n = \(\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(2 k^{3}+k^{2}\right)=2 \sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} k^{2}\)

Class 11 Maths NCERT Miscellaneous Solutions

Miscellaneous Exercise Class 11 Maths Question 2: If the sum of three numbers in AP is 24 and their product is 440, find the numbers. Ans: Let the three numbers in A.P. be a – d, a, and a + d. According to the given information, (a – d) + (a) + (a + d) = 24 3a = 24 ⇒ a = 8 and (a – d) a (a + d) = 440 ………………(ii) (8 – d) (8) (8 + d) = 440 (8 – d) (8 + d) = 55 = 64 – d 2 = 55 d 2 = 64 – 55 = 9 d = ±3 Therefore, when d = 3, the numbers are 5, 8 and 11 and when d = – 3, the numbers are 11, 8 and 5. Thus, the three numbers are 5, 8 and 11.

Miscellaneous Exercise Class 11 Maths Question 6: Find the sum of all two digit numbers which when divided by 4, yields 1 as remninder. Ans: The sum of two digit numbers divisible by 4 yield 1 as remainder is 13 + 17 + 21 + ………… + 97. Let the sum be denoted by S and let 97 be the nth term. ∴ T n = a + (n – 1) d 97 = a + (n – 1) d = 13 + (n – 1) 4 ⇒ 97 = 13 + 4n – 4 ⇒ 97 – 9 = 4n ⇒ n = 22 ∴ The sum, S n = 13 + 17 + 21 + ………….+ 97 ∴ S n = \(\frac{n}{2}\) [2a + (n – 1)d] = \(\frac{22}{7}\) [2 × 13 + (22 – 1) × 4] = 11 [26 + 21 × 4] = 11 [26 + 84] = 11 × 110 = 1210

Miscellaneous Exercise Class 11 Maths Question 8: The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. Ans: Let the sum of n terms of the G.P. be 315. It is known that, S n = \(\frac{a\left(r^{n}-1\right)}{r-1}\) It is given that the first term a is 5 and common ratio r is 2. 315 = \(\frac{5\left(2^{n}-1\right)}{2-1}\) ⇒ 2n = 64 = (2) 6 ⇒ n = 6 ∴ Last term of the G.P.= 6th term = ar 6 – 1 = (5) (2) 5 = (5) (32) = 160 Thus, the last term of the G.P. is 160.

Miscellaneous Exercise Class 11 Maths Question 10: The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. Ans: Let the three numbers in G.P. be a, ar and ar 2 . From the given condition, a + ar + ar 2 = 56 a (1 + r + r 2 ) = 56 a = \(\frac{56}{1+r+r^{2}}\) …………………(i) a – 1, ar – 7, ar 2 – 21 forms an A.P. ∴ (ar – 7) – (a – 1) = (ar 2 – 21) – (ar – 7) ⇒ ar – a – 6 = ar 2 – ar – 14 ⇒ ar 2 – 2ar + a = 8 ⇒ ar 2 – ar – ar + a = 8 a(r 2 + 1 – 2r) = 8 a(r – 1) 2 = 8 ………….(ii) \(\frac{56}{1+r+r^{2}}\) (r – 1) 2 = 8 [using eq. (1)] ⇒ 7 (r 2 – 2r + 1) = 1 + r + r 2 ⇒ 7r 2 – 14r + 7 – 1 – r – r 2 = 0 ⇒ 6r 2 – 15r + 6 = 0 ⇒ r 2 – 12r – 3r + 6 = 0 ⇒ 6r (r – 2) – 3 (r – 2) = 0 ⇒ (6r – 3)(r – 2) = 0 ∴ r = 2, \(\frac{1}{2}\) When r = 2, a = 8; When r = \(\frac{1}{2}\), a = 32 Therefore, when r = 2, the three numbers in G.P. are 8, 16 and 32. When r = \(\frac{1}{2}\), the three numbers in G.P. are 32, 16 and 8. Thus, in either case, the three required numbers are 8, 16, and 32.

Miscellaneous Exercise Class 11 Maths Question 15: The th, qth and rth terms of an AP. are a, b, c respectively. Show that(q – r)a + (r – p) b + (p – q) c = 0. Ans: Let A be the first term and d be the common difference. Since, T p = a ⇒ A + (p – 1) d = a ……………(i) T q = b ⇒ A + (q – 1) d = b …………….(ii) and T r = c ⇒ A + (r – 1) d = c ……………..(iii) (i) On multiplying eq. (i) by (q – r), eq. (ii) by (r – p) and eq. (iii) by (p – q),we get (q – r) A + (p – 1) (q – r) d = a (q – r) …………..(iv) (r – p) A + (q – 1) (r – p) d = b (r – p) …………….(v) and (p – q) A + (r – 1) (p – q) d = c (p – q) ……………(vi) On adding eqs. (iv), (v) and eq. (vi), we get (q – r) A + (p – 1) (q – r) d + (r – p) A + (q – 1) (r – p) d + (p – q) A + (r – 1) (p – q) d = a (q – r) + b (r – p) + c (p – q) ⇒ A [(q – r) + (r – p) + (p – q)] + (p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)] d = a (q – r) + b (r – p) + c (p – q) A(0) + (0)d = a (q – r) + b (r – p) + c (p – q) a (q – r) + b (r – p) + c (p – q) = 0 Hence proved.

Miscellaneous Exercise Class 11 Maths Question 16: If \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)\) are in A.P.,prove that a, b, c are in A.P. Ans:

It is given that \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)\) are in A.P.

∴ \(b\left(\frac{1}{c}+\frac{1}{a}\right)-a\left(\frac{1}{b}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)-b\left(\frac{1}{c}+\frac{1}{a}\right)\)

⇒ \(\frac{b(a+c)}{a c}-\frac{a(b+c)}{b c}=\frac{c(a+b)}{a b}-\frac{b(a+c)}{a c}\)

⇒ \(\frac{b^{2} a+b^{2} c-a^{2} b-a^{2} c}{a b c}=\frac{c^{2} a+c^{2} b-b^{2} a-b^{2} c}{a b c}\)

⇒ b 2 a – a 2 b + b 2 c – a 2 c = c 2 a – b 2 a + c 2 b – b 2 c ⇒ ab (b – a) + c (b 2 – a 2 ) = a (c 2 – b 2 ) + bc (c – b) ⇒ ab (b – a) + c (b – a) (b + a) = a (c – b) (c + b) + bc (c – b) ⇒ (b – a) (ab + cb + ca) = (c – b) (ac + ab + bc) ⇒ b – a = c – b Thus, a, b and c are in A.P.

NCERT Solutions for Class 11 Maths All Chapters

• Chapter 1 Sets
• Chapter 2 Relations and Functions
• Chapter 3 Trigonometric Functions
• Chapter 4 Principle of Mathematical Induction
• Chapter 5 Complex Numbers and Quadratic Equations
• Chapter 6 Linear Inequalities
• Chapter 7 Permutation and Combinations
• Chapter 8 Binomial Theorem
• Chapter 9 Sequences and Series
• Chapter 10 Straight Lines
• Chapter 11 Conic Sections
• Chapter 12 Introduction to Three Dimensional Geometry
• Chapter 13 Limits and Derivatives
• Chapter 14 Mathematical Reasoning
• Chapter 15 Statistics
• Chapter 16 Probability

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NCERT solutions for Mathematics Class 11 chapter 9 - Sequences and Series [Latest edition]

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Solutions for chapter 9: sequences and series.

Below listed, you can find solutions for Chapter 9 of CBSE, Karnataka Board PUC NCERT for Mathematics Class 11.

NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Exercise 9.1 [Pages 180 - 181]

Write the first five terms of the sequences whose n th term is:

`a_n = n(n+2)`

`a_n = n/(n + 1)`

a n  = 2 n

Write the first five terms of the sequences whose n th term is:

`a_n = (2n -3)/6`

`a_n = (-1)^(n-1) 5^(n+1)`

`a_n = n (n^2 + 5)/4`

Find the indicated term in the following sequence whose n th term is:

a n = 4n – 3; a 17 , a 24 

Find the indicated term in the following sequence whose n th  term is:

`a_n = n^2/2^n`; `a_7`

`a_n = (–1)^(n – 1) n^3; a_9`

`a_n = (n(n-2))/(n+3)` ;`a_20`

Write the first five terms of the following sequence and obtain the corresponding series:

a 1 = 3, a n = 3a (n - 1) + 2 for all n > 1

`a_1 = -1, a_n = (a_(n-1))/n , n >= 2`

Write the first five terms of the following sequence and obtain the corresponding series: 

`a_1 = a_2 = 2, a_n = a_(n-1) -1, n > 2`

The Fibonacci sequence is defined by 1 = a 1 = a 2 and an = a n – 1 + a n – 2 , n > 2.

Find `a_(n+1)/a_n`, for n = 1, 2, 3, 4, 5

NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Exercise 9.2 [Pages 185 - 186]

Find the sum of odd integers from 1 to 2001.

Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20 th  term is –112.

How many terms of the A.P.  -6 , `-11/2` , -5... are needed to give the sum –25?

In an A.P., if  p th  term is 1/q and qth term is 1/p,  prove that the sum of first  pq  terms is 1/2 (pq + 1) where `p != q`

If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term

Find the sum to  n  terms of the A.P., whose  k th  term is 5 k  + 1.

If the sum of  n  terms of an A.P. is ( pn  +  qn 2 ), where  p  and  q  are constants, find the common difference.

The sums of  n  terms of two arithmetic progressions are in the ratio 5 n  + 4: 9 n  + 6. Find the ratio of their 18 th  terms

If the sum of first  p  terms of an A.P. is equal to the sum of the first  q  terms, then find the sum of the first ( p  +  q ) terms.

Sum of the first  p, q  and  r  terms of an A.P. are  a, b  and  c , respectively.

Prove that `a/p (q - r) + b/q (r- p) + c/r (p - q) = 0`

The ratio of the sums of  m  and  n  terms of an A.P. is  m 2 :  n 2 . Show that the ratio of  m th  and  n th term is (2 m  – 1): (2 n  – 1)

If the sum of  n  terms of an A.P. is 3n 2 + 5n and its   m th   term is 164, find the value of  m .

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

if `(a^n + b^n)/(a^(n-1) + b^(n-1))` is the A.M. between  a  and  b , then find the value of  n .

Between 1 and 31,  m  numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7 th  and ( m  – 1) th  numbers is 5:9. Find the value of  m .

A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs 5 every month, what amount he will pay in the 30 th  installment?

The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Exercise 9.3 [Pages 192 - 193]

Find the 20 th  and n th terms of the G.P. `5/2, 5/4 , 5/8,...`

Find the 12 th  term of a G.P. whose 8 th  term is 192 and the common ratio is 2.

The 5 th , 8 th  and 11 th  terms of a G.P. are p, q and s, respectively. Show that q 2  = ps.

The 4 th  term of a G.P. is square of its second term, and the first term is –3. Determine its 7 th term.

Which term of the following sequence: 

`2, 2sqrt2, 4,.... is 128`

Which term of the following sequence:

`sqrt3, 3, 3sqrt3`, .... is 729?

`1/3, 1/9, 1/27`, ...., is `1/19683`?

For what values of x, the numbers  `-2/7, x, -7/2` are in G.P?

Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015,…

Find the sum to indicated number of terms of the geometric progressions `sqrt7, sqrt21,3sqrt7`...n terms.

Find the sum to indicated number of terms in the geometric progressions 1, – a, a 2 , – a 3 , ... n terms (if a ≠ – 1).

Find the sum to indicated number of terms in the geometric progressions x 3 , x 5 , x 7 , ... n terms (if x ≠ ± 1).

Evaluate `sum_(k=1)^11 (2+3^k )`

The sum of first three terms of a G.P. is  `39/10` and their product is 1. Find the common ratio and the terms.

How many terms of G.P. 3, 3 2 , 3 3 , … are needed to give the sum 120?

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Given a G.P. with a = 729 and 7 th  term 64, determine S 7 .

Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.

If the 4 th , 10 th  and 16 th  terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

Find the sum of the products of the corresponding terms of the sequences `2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2`

Show that the products of the corresponding terms of the sequences a, ar, ar 2 , …ar n – 1 and A, AR, AR 2 , … `AR^(n-1)`   form a G.P, and find the common ratio

Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4 th  by 18.

If the p th , q th and r th terms of a G.P. are a, b and c, respectively. Prove that `a^(q - r) b^(r-p) c^(p-q) = 1`

If the first and the n th  term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P 2  = (ab) n .

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1) th to (2n) th term is `1/r^n`.

If a, b, c and d are in G.P. show that (a 2 + b 2 + c 2 ) (b 2 + c 2 + d 2 ) = (ab + bc + cd) 2 .

Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Find the value of n so that  `(a^(n+1) + b^(n+1))/(a^n + b^n)` may be the geometric mean between a and b.

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio `(3 + 2sqrt2) ":" (3 - 2sqrt2)`.

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are `A+- sqrt((A+G)(A-G))`.

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2 nd  hour, 4 th  hour and n th hour? 

What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Exercise 9.4 [Page 196]

Find the sum to  n  terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Find the sum to  n  terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Find the sum to  n  terms of the series 3 × 1 2  + 5 × 2 2  + 7 × 3 2  + …

Find the sum to  n  terms of the series `1/(1xx2) + 1/(2xx3)+1/(3xx4)+ ...`

Find the sum to  n  terms of the series  5 2 + 6 2 + 7 2 + ... + 20 2

Find the sum to  n  terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Find the sum to  n  terms of the series 1 2  + (1 2  + 2 2 ) + (1 2  + 2 2  + 3 2 ) + …

Find the sum to  n  terms of the series whose  n th  term is given by  n  ( n  + 1) ( n  + 4).

Find the sum to n terms of the series whose n th  terms is given by n 2  + 2 n

Find the sum to  n  terms of the series whose  n th  terms is given by (2 n  – 1) 2

NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Miscellaneous Exercise [Pages 199 - 200]

Show that the sum of ( m  +  n ) th  and ( m  –  n ) th  terms of an A.P. is equal to twice the  m t h  term.

If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Let the sum of  n , 2 n , 3 n  terms of an A.P. be S 1 , S 2  and S 3 ,   respectively, show that S 3  = 3 (S 2 – S 1 )

Find the sum of all numbers between 200 and 400 which are divisible by 7.

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and `sum_(x = 1)^n` f(x) = 120, find the value of n.

The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

if `(a+ bx)/(a - bx) = (b +cx)/(b - cx) = (c + dx)/(c- dx) (x != 0)` then show that a, b, c and d are in G.P.

Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P 2 R n   = S n

The  p th ,  q th  and  r th  terms of an A.P. are  a, b, c  respectively. Show that (q – r )a + (r – p )b + (p – q )c = 0

if `a(1/b + 1/c), b(1/c+1/a), c(1/a+1/b)` are in A.P., prove that  a, b, c  are in A.P.

If a, b, c, d are in G.P, prove that (a n + b n ), (b n + c n ), (c n + d n ) are in G.P.

If a and b are the roots of are roots of x 2 – 3x + p = 0 , and c, d are roots of x 2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17 : 15.

The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that `a:b = (m + sqrt(m^2 - n^2)):(m - sqrt(m^2 - n^2))`.

If  a, b, c  are in A.P , ;  b, c, d  are in G.P and ` 1/c, 1/d,1/e` are in A.P. prove that  a ,  c ,  e  are in G.P.

Find the sum of the following series up to n terms:

5 + 55 + 555 + …

.6 +.66 +. 666 +…

Find the 20 th  term of the series 2 × 4 + 4 × 6 + 6 × 8 + … +  n  terms.

Find the sum of the first  n  terms of the series: 3 + 7 + 13 + 21 + 31 + …

If S 1 , S 2 , S 3  are the sum of first  n  natural numbers, their squares and their cubes, respectively, show that  `9S_2^2 = S_3(1 + 8S_1)`

Find the sum of the following series up to  n  terms `1^3/1 + (1^3 + 2^3)/(1+3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) +...`

Show that  `(1xx2^2 + 2xx3^2 + ...+nxx(n+1)^2)/(1^2 xx 2 + 2^2 xx3 + ... + n^2xx (n+1))` = `(3n + 5)/(3n + 1)`

A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installments of Rs 500 plus 12% interest on the unpaid amount. How much will be the tractor cost him?

Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8 th  set of letter is mailed.

A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15 th  year since he deposited the amount and also calculate the total amount after 20 years.

A manufacturer reckons that the value of a machine, which costs him Rs 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

NCERT solutions for Mathematics Class 11 chapter 9 - Sequences and Series

Shaalaa.com has the CBSE, Karnataka Board PUC Mathematics Mathematics Class 11 CBSE, Karnataka Board PUC solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT solutions for Mathematics Mathematics Class 11 CBSE, Karnataka Board PUC 9 (Sequences and Series) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Mathematics Class 11 chapter 9 Sequences and Series are Sum to N Terms of Special Series, Introduction of Sequence and Series, Concept of Sequences, Concept of Series, Arithmetic Progression (A.P.), Geometric Progression (G. P.), Relationship Between A.M. and G.M., Sum to N Terms of Special Series, Introduction of Sequence and Series, Concept of Sequences, Concept of Series, Arithmetic Progression (A.P.), Geometric Progression (G. P.), Relationship Between A.M. and G.M..

Using NCERT Mathematics Class 11 solutions Sequences and Series exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Solutions are essential questions that can be asked in the final exam. Maximum CBSE, Karnataka Board PUC Mathematics Class 11 students prefer NCERT Textbook Solutions to score more in exams.

Get the free view of Chapter 9, Sequences and Series Mathematics Class 11 additional questions for Mathematics Mathematics Class 11 CBSE, Karnataka Board PUC, and you can use Shaalaa.com to keep it handy for your exam preparation.

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NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series

Ncert solutions for class 11 maths chapter 9 – sequences and series pdf.

Free PDF of NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. To download our free pdf of Chapter 9 – Sequences and Series Maths NCERT Solutions for Class 11 to help you to score more marks in your board exams and as well as competitive exams.

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Sequence and Series Word Problems | Class 11 Maths

Sequences and series have several important applications in several spheres of human activities. When sequences follow some specific patterns, they are usually called progressions. Arithmetic and Geometric progressions are some examples of commonly occurring progressions. Let’s see some problems on these progressions to understand them better. 

Growth Patterns

Such types of number sequence problems first describe how a sequence of numbers is generated. Some terms of the sequence are given, and we need to figure out the patterns in them and then the next terms of the sequence. 

Solving such sequences: 

• Look for a pattern between the given numbers.
• Decide whether to use +, -, × or ÷
• Use the pattern to solve the sequence.

Increasing Pattern

Increasing Patterns as the name suggests will always be increasing in nature. The next term and the term just before that will all be similarly related with the help of operations (×, -, +).

Question: 6,13,27,55, ….. In the given sequence, what is the value of the next term? 

On looking carefully into the pattern, one can see that 13 = 6 × 2 + 1  27 = 13 × 2 + 1  55 = 27 × 2 + 1  This shows that every term is twice the preceding term plus one. So, let the next term be “a”.  a = 55 × 2 + 1     = 110 + 1     = 111  Hence, the next term is 111. 

Decreasing Pattern

In decreasing Patterns, the next will be lesser than the previous term and two consecutive terms will follow a certain pattern.

Question: What is the next term in the series: 220,100,40, ….

The series follow the pattern: 100 = (220×0.5)-10                                                 40 = (100×0.5)-10 Therefore, the next term will be (40×0.5)-10 = 10

Arithmetic Progressions 

In Arithmetic Progressions , the consecutive terms will have the same difference and are denoted as ‘d’, the first term is called as ‘a’, and the number of terms is denoted as ‘n’. 

The formula for n th term is: T n = a+ (n-1)d

Question 1: 2, 5, 8, 11, …. Find the next term of the sequence. 

It can be noticed by carefully studying the terms of the sequence that the difference between each consecutive term remains the same. For example:  5 – 2 = 3  8 – 5 = 3 11 – 8 = 3  So, the next will be at a difference of three from the last term. Since the last term of the sequence is 11. The next terms will be 14. OR The formula for n th term in Arithmetic progression can also be used here, 5 th term is required here: a = 2 d = (3) T 5 = 2+(5-1)(3)  = 2+12 = 14

Question 2: 15, 12, 9, … __. Find the next term.

In this problem also, all the terms have a difference of three between them. The difference is that the sequence in decreasing in nature. Since the last term is 9, the next term will be 3 less than the last term. So, the last term will be 6. 

Fibonacci Sequence

Sometimes there are sequences for which pattern is not visible, the Fibonacci sequence is an example of such a sequence. It is a very commonly occurring sequence in the field of mathematics and computer science.

The number is arranged as, 1, 1, 2, 3, 5, 8 …. Here the pattern is not visible, this sequence proceeds in a manner that depends on its history. 

Let a n be the nth term for the sequence. In this sequence a n = a n-1 + a n-2 . 

For example, 

a 2 = a 1 + a 0 i.e 2 = 1 + 1,  a 3 = a 1 + a 2 i.e 3 = 2 + 1,  a 4 = a 3 + a 2 i.e 5 = 2 + 3 and,  a 5 = a 4 + a 3 i.e 8 = 5 + 3. 

Question: What is the next term in the Fibonacci series: 1,1,2,3,5,8,13,……

In Fibonacci Series, the next term is the sum of the last two terms. Therefore, the next term will be (8+13)= 21

Geometric Series

Before starting out with the problems related to a geometric progression. Let’s take a quick recap of the formulas for the sum and nth term of a GP. 

The general form of a geometric progression is a, ar, ar 2 , ar 3 …… where a = first term, r = common ratio and a n be the nth term. 

• The nth term of the progression: a n = ar n-1 .

Finite GP problems

These kinds of problems include the geometric series where there are a finite number of terms. 

Question 1: The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour, and nth hour? 

The growth of the bacteria makes a GP, 30, 60, 120,….. and so on.  In this GP, a = 30, r = 2. Let the number of bacteria at nth hour be a n .  At n = 2, a 2 = ar 2-1                       = (30)(2) 2-1 = 30 × 2 = 60  At n = 4, a 4 = ar 4-1                       = (30)(2) 4-1 = 30 × 2 3 = 240 At nth step an = ar (n-1) = 30 × 2 n-1

Question 2: A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own. 

This is a problem of finite GP. The sequence can be thought like this,  2, 4, 8, 16, …..  So, the total number of ancestors in 10 generation of his family.  2, 4, 8, 16, 32, …..10 terms.  Here, a = 2, r = 2 and n = 10  S10 = a 

Infinite GP Problems

An infinite geometric series is the sum of an infinite geometric sequence. This series would have no last term

Question 1: A monkey is swinging from a tree. On the first swing, she passes through an arc of 24m. With each swing, she passes through an arc of 24m. With each swing, she passes through an arc half the length of the previous swing. What is the total distance traveled by the monkey when she completes her 100000th swing? 

Now this movement represents a GP with a = 24 and r = 1/2. Now, since the GP is decreasing and the question asks for the sum till 100000th term. To save the calculation , we can consider it an infinite GP and round of the answer we get.  Sum of an infinite GP =  Here, a = 24 and r = 1/2. Let the sum be S  So the monkey travels almost 24m in these many swings. 

Question 2: A ball was dropped from a 24-inch high table. The ball rebounds and always reaches three fourth of the distance fallen. What is the approximate distance the ball travels before finally coming at rest on the ground. 

It should be noticed that this problem actually involves two infinite geometric series. The first series involves ball falling and the other series involves ball rising after rebounding from the ground.  Falling: a 1 = 24 , r = 3/4  Rising: a 2 = 24(3/4) = 18 ,r = 3/4  Using the formula for infinite geometric series,  S =  Let S be the total distance travelled: S = S rising + S falling S rising = S falling =  Now, S = S rising + S falling = 72 + 96 = 168

Practice Problems – Sequence and Series Word Problems

• Find the next term in the sequence: 3, 7, 15, 31…..?
• Determine the next term in the sequence: 120, 80, 60, 48…..?
• Find the next term in the sequence: 7, 14, 21, 28,…..?
• What is the 10th term in the arithmetic progression: 4, 9, 14, 19…..?
• Find the next term in the sequence: 1, 1, 2, 3, 5, 8, 13, 21,….?
• A bacteria culture starts with 50 bacteria and triples every hour. How many bacteria will there be after 5 hours?

FAQs – Sequence and Series Word Problems

What is the difference between an arithmetic progression and a geometric progression.

In an arithmetic progression (AP), the difference between consecutive terms is constant. In a geometric progression (GP), the ratio between consecutive terms is constant.

How do you identify the pattern in a sequence?

To identify the pattern, look for consistent operations (addition, subtraction, multiplication, division) applied to the terms. For complex sequences, consider recursive relationships or consult specific formulas.

What is the sum of an arithmetic series?

The sum of the first n terms of an arithmetic series can be calculated using the formula: where a is the first term, d is the common difference and n is the number of terms.

How do you find the sum of a finite geometric series?

The sum of the first n terms of a finite geometric series is given by: where a is the first term, r is the common ratio, and n is the number of terms.

What is the significance of the Fibonacci sequence in real life?

The Fibonacci sequence appears in various natural phenomena, including the arrangement of leaves on a stem, the branching of trees, and the arrangement of a pine cone’s scales. It is also used in algorithms, financial models, and art due to its unique properties.

Similar Reads

• Sequences and Series Practice Problems
• Sequences and Series Formulas
• Practice Questions on Arithmetic Progression
• Class 11 NCERT Solutions- Chapter 9 Sequences And Series – Exercise 9.1
• Class 11 NCERT Solutions- Chapter 9 Sequences And Series – Exercise 9.2
• NCERT Solutions Class 11 – Chapter 8 Sequences And Series – Exercise 8.1

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Download Case Study Questions for Class 11 Maths

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[PDF] Download Case Study Questions for Class 11 Maths

Here we are providing case study questions for Class 11 Maths. In this article, we are sharing links for Class 11 Maths All Chapters. All case study questions of Class 11 Maths are solved so that students can check their solutions after attempting questions.

Click on the chapter to view.

Class 11 Maths Chapters

Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutation and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight Lines Chapter 11 Conic Sections Chapter 12 Introduction to Three-Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics Chapter 16 Probability

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills. Here are some reasons why case study questions are important in Class 11 maths education:

• Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
• Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
• Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
• Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
• Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

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NCERT Solutions for Class 11 Maths Chapter 8 - Sequences and Series

NCERT Chapter 8 Maths Sequences and Series Class 11 Solutions - Free PDF Download

The NCERT Maths Chapter 8 Sequence and Series Class 11 Solutions is all about understanding the order and pattern of numbers. The free PDF of Chapter 8 Class 11 Sequence and Series Maths Solutions is available on Vedantu, providing students with a better understanding of the problems. It covers solutions to every exercise in this chapter and is updated according to the latest CBSE syllabus. In this chapter, you will learn about different types of sequences and geometric sequences. Vedantu’s solutions provide step-by-step explanations to help you grasp these concepts easily. The clear and detailed solutions ensure that you understand how to approach and solve problems related to sequences and series.

Glance on Class 11th Maths Chapter 8 - Sequences and Series

Sequence, we mean an arrangement of numbers in definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type $\{1,2,3, \ldots i k$. 

A sequence containing a finite number of terms is called a finite sequence A sequence is called infinite if it is not a finite sequence

Let $a_1 a_2, a_3, \ldots$ be the sequence, then the sum expressed as $a_1+a_2+a_3+\ldots$. is called a series. A series is called finite series ifit has got finite number of terms

This article contains chapter notes, important questions, exemplar solutions and exercises links for Chapter 8 - Sequences and Series, which you can download as PDFs.

There are two exercises (46 fully solved questions) and one Miscellaneous exercise(18 fully solved questions) in class 11th Maths Chapter 8 Sequences and Series.

Access Exercise wise NCERT Solutions for Chapter 8 Maths Class 11

Current Syllabus Exercises of Class 11 Maths Chapter 8

NCERT Solutions of Class 11 Maths Sequences and Series Miscellaneous Exercise

Exercises Under NCERT Solutions for Class 11 Maths Chapter 8 – Sequences and Series

Exercise 8.1: This exercise contains 14 fully solved questions. This exercise introduces the concept of sequences and their types, including arithmetic sequences, geometric sequences, and harmonic sequences. Students will practice identifying the nth term of each type of sequence.

Exercise 8.2: This exercise contains 32 fully solved questions. This exercise focuses on Geometric Progression (GP) and its various properties. Students will learn about the nth term and the sum of n terms of a GP and how to apply these concepts in problem-solving.

Miscellaneous Exercise: This exercise contains 18 fully solved questions. This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of sequences and series to solve various problems and answer questions.

Access NCERT Solutions for Class 11 Maths Chapter 8 – Sequences and Series

Exercise 8.1.

1. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}=n\left( n+2 \right)\] .

The given equation is \[{{a}_{n}}=n\left( n+2 \right)\] .

Substitute \[n=1\] in the equation.

\[{{a}_{1}}=1\left( 1+2 \right)\]

\[\Rightarrow {{a}_{1}}=3\]

Similarly substitute \[n=2,3,4\] and \[5\]in the equation.

\[{{a}_{2}}=2\left( 2+2 \right)\]

\[\Rightarrow {{a}_{2}}=8\]

\[{{a}_{3}}=3\left( 3+2 \right)\]

\[\Rightarrow {{a}_{3}}=15\]

\[{{a}_{4}}=4\left( 4+2 \right)\]

\[\Rightarrow {{a}_{4}}=24\]

\[{{a}_{5}}=5\left( 5+2 \right)\]

\[\Rightarrow {{a}_{5}}=35\]

Therefore, the first five terms of \[{{a}_{n}}=n\left( n+2 \right)\] is \[3,8,15,24\] and \[35\] . 

2. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}=\frac{n}{n+1}\] .

The given equation is \[{{a}_{n}}=\frac{n}{n+1}\] .

\[{{a}_{1}}=\frac{1}{1+1}\]

\[\Rightarrow {{a}_{1}}=\frac{1}{2}\]

\[{{a}_{2}}=\frac{2}{2+1}\]

\[\Rightarrow {{a}_{2}}=\frac{2}{3}\]

\[{{a}_{3}}=\frac{3}{3+1}\]

\[\Rightarrow {{a}_{3}}=\frac{3}{4}\]

\[{{a}_{4}}=\frac{4}{4+1}\]

\[\Rightarrow {{a}_{4}}=\frac{4}{5}\]

\[{{a}_{5}}=\frac{5}{5+1}\]

\[\Rightarrow {{a}_{5}}=\frac{5}{6}\]

Therefore, the first five terms of \[{{a}_{n}}=\frac{n}{n+1}\] is \[\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}\] and \[\frac{5}{6}\] . 

3. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}={{2}^{n}}\] .

The given equation is \[{{a}_{n}}={{2}^{n}}\] .

\[{{a}_{1}}={{2}^{1}}\]

\[\Rightarrow {{a}_{1}}=2\]

\[{{a}_{2}}={{2}^{2}}\]

\[\Rightarrow {{a}_{2}}=4\]

\[{{a}_{3}}={{2}^{3}}\]

\[\Rightarrow {{a}_{3}}=8\]

\[{{a}_{4}}={{2}^{4}}\]

\[\Rightarrow {{a}_{4}}=16\]

\[{{a}_{5}}={{2}^{5}}\]

\[\Rightarrow {{a}_{5}}=32\]

Therefore, the first five terms of \[{{a}_{n}}={{2}^{n}}\] is \[2,4,8,16\] and \[32\] . 

4. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}=\frac{2n-3}{6}\] .

The given equation is \[{{a}_{n}}=\frac{2n-3}{6}\] .

\[{{a}_{1}}=\frac{2\left( 1 \right)-3}{6}\]

\[\Rightarrow {{a}_{1}}=-\frac{1}{6}\]

\[{{a}_{2}}=\frac{2\left( 2 \right)-3}{6}\]

\[\Rightarrow {{a}_{2}}=\frac{1}{6}\]

\[{{a}_{3}}=\frac{2\left( 3 \right)-3}{6}\]

\[\Rightarrow {{a}_{3}}=\frac{3}{6}=\frac{1}{2}\]

\[{{a}_{4}}=\frac{2\left( 4 \right)-3}{6}\]

\[\Rightarrow {{a}_{4}}=\frac{5}{6}\]

\[{{a}_{5}}=\frac{2\left( 5 \right)-3}{6}\]

\[\Rightarrow {{a}_{5}}=\frac{7}{6}\]

Therefore, the first five terms of \[{{a}_{n}}=\frac{2n-3}{6}\] is \[-\frac{1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6}\] and \[\frac{7}{6}\] . 

5. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{5}^{n+1}}\] .

The given equation is \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{5}^{n+1}}\] .

\[{{a}_{1}}={{\left( -1 \right)}^{1-1}}{{5}^{1+1}}\]

\[\Rightarrow {{a}_{1}}={{5}^{2}}=25\]

\[{{a}_{2}}={{\left( -1 \right)}^{2-1}}{{5}^{2+1}}\]

\[\Rightarrow {{a}_{2}}=-{{5}^{3}}=-125\]

\[{{a}_{3}}={{\left( -1 \right)}^{3-1}}{{5}^{3+1}}\]

\[\Rightarrow {{a}_{3}}={{5}^{4}}=625\]

\[{{a}_{4}}={{\left( -1 \right)}^{4-1}}{{5}^{4+1}}\]

\[\Rightarrow {{a}_{4}}=-{{5}^{5}}=-3125\]

\[{{a}_{5}}={{\left( -1 \right)}^{5-1}}{{5}^{5+1}}\]

\[\Rightarrow {{a}_{5}}={{5}^{6}}=15625\]

Therefore, the first five terms of \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{5}^{n+1}}\] is \[25,-125,625,-3125\] and \[15625\] . 

6. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}=n\frac{{{n}^{2}}+5}{4}\] .

The given equation is \[{{a}_{n}}=n\frac{{{n}^{2}}+5}{4}\] .

\[{{a}_{1}}=1\cdot \frac{{{1}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{1}}=\frac{6}{4}=\frac{3}{2}\]

\[{{a}_{2}}=2\cdot \frac{{{2}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{2}}=\frac{18}{4}=\frac{9}{2}\]

\[{{a}_{3}}=3\cdot \frac{{{3}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{3}}=\frac{42}{4}=\frac{21}{2}\]

\[{{a}_{4}}=4\cdot \frac{{{4}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{4}}=\frac{84}{4}=21\]

\[{{a}_{5}}=5\cdot \frac{{{5}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{5}}=\frac{150}{4}=\frac{75}{2}\]

Therefore, the first five terms of \[{{a}_{n}}=n\frac{{{n}^{2}}+5}{4}\] is \[\frac{3}{2},\frac{9}{2},\frac{21}{2},21\] and \[\frac{75}{2}\] . 

7. Find the \[{{17}^{th}}\] and \[{{24}^{th}}\] term in the following sequence whose \[{{n}^{th}}\] term is \[{{a}_{n}}=4n-3\] .

The given equation is \[{{a}_{n}}=4n-3\] .

Substitute \[n=17\] in the equation.

\[{{a}_{17}}=4\left( 17 \right)-3\]

\[\Rightarrow {{a}_{17}}=65\]

Similarly substitute \[n=24\] in the equation.

\[{{a}_{24}}=4\left( 24 \right)-3\]

\[\Rightarrow {{a}_{24}}=93\]

Therefore, the \[{{17}^{th}}\] and \[{{24}^{th}}\] term of \[{{a}_{n}}=4n-3\] is \[65\] and \[93\] respectively. 

8. Find the \[{{7}^{th}}\] term in the following sequence whose \[{{n}^{th}}\] term is \[{{a}_{n}}=\frac{{{n}^{2}}}{{{2}^{n}}}\] .

The given equation is \[{{a}_{n}}=\frac{{{n}^{2}}}{{{2}^{n}}}\] .

Substitute \[n=7\] in the equation.

\[{{a}_{7}}=\frac{{{7}^{2}}}{{{2}^{7}}}\]

\[\Rightarrow {{a}_{7}}=\frac{49}{128}\]

Therefore, the \[{{7}^{th}}\] term of \[{{a}_{n}}=\frac{{{n}^{2}}}{{{2}^{n}}}\] is \[\frac{49}{128}\] . 

9. Find the \[{{9}^{th}}\] term in the following sequence whose \[{{n}^{th}}\] term is \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{n}^{3}}\] .

The given equation is \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{n}^{3}}\] .

Substitute \[n=9\] in the equation.

\[{{a}_{9}}={{\left( -1 \right)}^{9-1}}{{9}^{3}}\]

\[\Rightarrow {{a}_{9}}=729\]

Therefore, the \[{{9}^{th}}\] term of \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{n}^{3}}\] is \[729\] . 

10. Find the \[{{20}^{th}}\] term in the following sequence whose \[{{n}^{th}}\] term is \[{{a}_{n}}=\frac{n\left( n-2 \right)}{n+3}\] .

The given equation is \[{{a}_{n}}=\frac{n\left( n-2 \right)}{n+3}\] .

Substitute \[n=20\] in the equation.

\[{{a}_{20}}=\frac{20\left( 20-2 \right)}{20+3}\]

\[\Rightarrow {{a}_{20}}=\frac{360}{23}\]

Therefore, the \[{{20}^{th}}\] term of \[{{a}_{n}}=\frac{n\left( n-2 \right)}{n+3}\] is \[\frac{360}{23}\] . 

11. Write the first five terms of the following sequence and obtain the corresponding series: \[{{a}_{1}}=3\] , \[{{a}_{n}}=3{{a}_{n-1}}+2\] for all \[n>1\] .

The given equation is \[{{a}_{n}}=3{{a}_{n-1}}+2\] where \[{{a}_{1}}=3\] and \[n>1\] .

Substitute \[n=2\] and \[{{a}_{1}}=3\] in the equation.

\[{{a}_{2}}=3{{a}_{2-1}}+2=3\left( 3 \right)+2\]

\[\Rightarrow {{a}_{2}}=11\]

Similarly substitute \[n=3,4\] and \[5\] in the equation.

\[{{a}_{3}}=3{{a}_{3-1}}+2=3\left( 11 \right)+2\]

\[\Rightarrow {{a}_{3}}=35\]

\[{{a}_{4}}=3{{a}_{4-1}}+2=3\left( 35 \right)+2\]

\[\Rightarrow {{a}_{4}}=107\]

\[{{a}_{5}}=3{{a}_{5-1}}+2=3\left( 107 \right)+2\]

\[\Rightarrow {{a}_{5}}=323\]

Therefore, the first five terms of \[{{a}_{n}}=3{{a}_{n-1}}+2\] is \[3,11,35,107\] and \[323\] .

The corresponding series obtained from the sequence is \[3+11+35+107+323+...\]

12. Write the first five terms of the following sequence and obtain the corresponding series: \[{{a}_{1}}=-1\] , \[{{a}_{n}}=\frac{{{a}_{n-1}}}{n}\] , \[n\ge 2\] .

The given equation is \[{{a}_{n}}=\frac{{{a}_{n-1}}}{n}\] where \[{{a}_{1}}=-1\] and \[n\ge 2\] .

Substitute \[n=2\] and \[{{a}_{1}}=-1\] in the equation.

\[{{a}_{2}}=\frac{{{a}_{2-1}}}{2}=\frac{-1}{2}\]

\[\Rightarrow {{a}_{2}}=-\frac{1}{2}\]

\[{{a}_{3}}=\frac{{{a}_{3-1}}}{3}=\frac{{}^{-1}/{}_{2}}{3}\]

\[\Rightarrow {{a}_{3}}=-\frac{1}{6}\]

\[{{a}_{4}}=\frac{{{a}_{4-1}}}{4}=\frac{{}^{-1}/{}_{6}}{4}\]

\[\Rightarrow {{a}_{4}}=-\frac{1}{24}\]

\[{{a}_{5}}=\frac{{{a}_{5-1}}}{5}=\frac{{}^{-1}/{}_{24}}{5}\]

\[\Rightarrow {{a}_{5}}=-\frac{1}{120}\]

Therefore, the first five terms of \[{{a}_{n}}=\frac{{{a}_{n-1}}}{n}\] is \[-1,-\frac{1}{2},-\frac{1}{6},-\frac{1}{24}\] and \[-\frac{1}{120}\] .

The corresponding series obtained from the sequence is \[\left( -1 \right)+\left( -\frac{1}{2} \right)+\left( -\frac{1}{6} \right)+\left( -\frac{1}{24} \right)+\left( -\frac{1}{120} \right)...\]

13. Write the first five terms of the following sequence and obtain the corresponding series: \[{{a}_{1}}={{a}_{2}}=2\] , \[{{a}_{n}}={{a}_{n-1}}-1\] , \[n>2\] .

The given equation is \[{{a}_{n}}={{a}_{n-1}}-1\] where \[{{a}_{1}}={{a}_{2}}=2\] and \[n>2\] .

Substitute \[n=3\] and \[{{a}_{2}}=2\] in the equation.

\[{{a}_{3}}={{a}_{3-1}}-1=2-1\]

\[\Rightarrow {{a}_{3}}=1\]

Similarly substitute \[n=4\] and \[5\] in the equation.

\[{{a}_{4}}={{a}_{4-1}}-1=1-1\]

\[\Rightarrow {{a}_{4}}=0\]

\[{{a}_{5}}={{a}_{5-1}}-1=0-1\]

\[\Rightarrow {{a}_{5}}=-1\]

Therefore, the first five terms of \[{{a}_{n}}={{a}_{n-1}}-1\] is \[2,2,1,0\] and \[-1\] .

The corresponding series obtained from the sequence is \[2+2+1+0+\left( -1 \right)+...\]

14. The Fibonacci sequence is defined by \[1={{a}_{1}}={{a}_{2}}\] , \[{{a}_{n}}={{a}_{n-1}}+{{a}_{n-2}}\] , \[n>2\] . Find \[\frac{{{a}_{n+1}}}{{{a}_{n}}}\], for \[n=1,2,3,4,5\] .

The given equation is \[{{a}_{n}}={{a}_{n-1}}+{{a}_{n-2}}\] where \[1={{a}_{1}}={{a}_{2}}\] and \[n>2\] .

Substitute \[n=3\] and \[1={{a}_{1}}={{a}_{2}}\] in the equation.

\[{{a}_{3}}={{a}_{3-1}}+{{a}_{3-2}}=1+1\]

\[\Rightarrow {{a}_{3}}=2\]

Similarly substitute \[n=4,5\] and \[6\] in the equation.

\[{{a}_{4}}={{a}_{4-1}}+{{a}_{4-2}}=2+1\]

\[\Rightarrow {{a}_{4}}=3\]

\[{{a}_{5}}={{a}_{5-1}}+{{a}_{5-2}}=3+2\]

\[\Rightarrow {{a}_{5}}=5\]

\[{{a}_{6}}={{a}_{6-1}}+{{a}_{6-2}}=5+3\]

\[\Rightarrow {{a}_{6}}=8\]

Substitute the values of  \[{{a}_{1}}\] and \[{{a}_{2}}\] in the expression \[\frac{{{a}_{n+1}}}{{{a}_{n}}}\] for \[n=1\] .

\[\Rightarrow \frac{{{a}_{1+1}}}{{{a}_{1}}}=\frac{1}{1}=1\]

Similarly, when \[n=2\],

\[\Rightarrow \frac{{{a}_{2+1}}}{{{a}_{2}}}=\frac{2}{1}=2\]

When \[n=3\],

\[\Rightarrow \frac{{{a}_{3+1}}}{{{a}_{3}}}=\frac{3}{2}\]

When \[n=4\],

\[\Rightarrow \frac{{{a}_{4+1}}}{{{a}_{4}}}=\frac{5}{3}\]

When \[n=5\],

\[\Rightarrow \frac{{{a}_{5+1}}}{{{a}_{5}}}=\frac{8}{5}\]

Therefore, the value of \[\frac{{{a}_{n+1}}}{{{a}_{n}}}\] for \[n=1,2,3,4,5\] is \[1,2,\frac{3}{2},\frac{5}{3}\] and \[\frac{8}{5}\] respectively.

Exercise 8.2

1. Find the \[{{20}^{th}}\] and \[{{n}^{th}}\] term of the G.P. \[\frac{5}{2},\frac{5}{4},\frac{5}{8},...\]

\[\frac{5}{2},\frac{5}{4},\frac{5}{8},...\] is the given G.P.

The first term of the G.P. is \[a=\frac{5}{2}\] and the common ratio is \[r=\frac{{5}/{4}\;}{{5}/{2}\;}=\frac{1}{2}\].

The \[{{n}^{th}}\] term of the G.P. is given by the equation \[{{a}_{n}}=a{{r}^{n-1}}\].

Substituting the values of \[a\] and \[r\] we get

\[{{a}_{n}}=\frac{5}{2}{{\left( \frac{1}{2} \right)}^{n-1}}=\frac{5}{\left( 2 \right){{\left( 2 \right)}^{n-1}}}=\frac{5}{{{\left( 2 \right)}^{n}}}\]

Similarly, the \[{{20}^{th}}\] term of the G.P. is \[{{a}_{20}}=a{{r}^{20-1}}\]

\[\Rightarrow {{a}_{20}}=\frac{5}{2}{{\left( \frac{1}{2} \right)}^{19}}=\frac{5}{\left( 2 \right){{\left( 2 \right)}^{19}}}=\frac{5}{{{\left( 2 \right)}^{20}}}\]

Therefore, the \[{{20}^{th}}\] and \[{{n}^{th}}\] term of the given G.P. is \[\frac{5}{{{\left( 2 \right)}^{20}}}\] and \[\frac{5}{{{\left( 2 \right)}^{n}}}\] respectively.

2. Find the \[{{12}^{th}}\] term of a G.P. whose \[{{8}^{th}}\] term is \[192\] and the common ratio is \[2\]. 

Let the first term of the G.P. be \[a\] and the common ratio \[r=2\] .

The \[{{8}^{th}}\] term of the G.P. is given by the equation \[{{a}_{8}}=a{{r}^{8-1}}\].

Substituting the values of \[{{a}_{8}}\] and \[r\] we get

\[\Rightarrow 192=a{{\left( 2 \right)}^{7}}\]

\[\Rightarrow {{\left( 2 \right)}^{6}}\left( 3 \right)=a{{\left( 2 \right)}^{7}}\]

\[\Rightarrow a=\frac{{{\left( 2 \right)}^{6}}\left( 3 \right)}{{{\left( 2 \right)}^{7}}}=\frac{3}{2}\]

Then \[{{12}^{th}}\] term of the G.P. is given by the equation \[{{a}_{12}}=a{{r}^{12-1}}\].

Substitute the values of \[a\] and \[r\] in the equation.

\[{{a}_{12}}=\frac{3}{2}{{\left( 2 \right)}^{11}}\]

\[=3{{\left( 2 \right)}^{10}}\]

Therefore, the \[{{12}^{th}}\] term of the G.P. is \[3072\] .

3. The \[{{5}^{th}}\], \[{{8}^{th}}\] and \[{{11}^{th}}\] terms of a G.P. are \[p\] ,\[q\] and \[s\] , respectively. Show that \[{{q}^{2}}=ps\] .  

Let the first term and the common ratio of the G.P. be \[a\] and \[r\] respectively.

According to the conditions given in the question,

\[{{a}_{5}}=a{{r}^{5-1}}=a{{r}^{4}}=p\]

\[{{a}_{8}}=a{{r}^{8-1}}=a{{r}^{7}}=q\]

\[{{a}_{11}}=a{{r}^{11-1}}=a{{r}^{10}}=s\]

Dividing \[{{a}_{8}}\] by \[{{a}_{5}}\] we get

\[\frac{a{{r}^{7}}}{a{{r}^{4}}}=\frac{q}{p}\]

\[\Rightarrow {{r}^{3}}=\frac{q}{p}\]

Dividing \[{{a}_{11}}\] by \[{{a}_{8}}\] we get

\[\frac{a{{r}^{10}}}{a{{r}^{7}}}=\frac{s}{q}\]

\[\Rightarrow {{r}^{3}}=\frac{s}{q}\]

Equate both the values of \[{{r}^{3}}\] obtained.

\[\frac{q}{p}=\frac{s}{q}\]

\[\Rightarrow {{q}^{2}}=ps\]

Therefore,  \[{{q}^{2}}=ps\] is proved.

4. The \[{{4}^{th}}\] term of a G.P. is square of its second term, and the first term is \[-3\] . Determine its \[{{7}^{th}}\] term.

It is given that \[a=-3\] .

\[{{a}_{4}}=a{{r}^{3}}=\left( -3 \right){{r}^{3}}\]

\[{{a}_{2}}=a{{r}^{1}}=\left( -3 \right)r\]

\[\left( -3 \right){{r}^{3}}={{\left[ \left( -3 \right)r \right]}^{2}}\]

\[\Rightarrow -3{{r}^{3}}=9{{r}^{2}}\]

\[\Rightarrow r=-3{{a}_{7}}\]

\[=a{{r}^{6}}\]

\[=\left( -3 \right){{\left( -3 \right)}^{6}}\]

\[=-{{\left( 3 \right)}^{7}}\]

Therefore, \[-2187\] is the seventh term of the G.P.

5. Which term of the following sequences:

\[2,2\sqrt{2},4...\] is \[128\] ?

\[2,2\sqrt{2},4...\] is the given sequence.

The first term of the G.P. \[a=2\] and the common ratio \[r={\left( 2\sqrt{2} \right)}/{2}\;=\sqrt{2}\] .

\[128\] is the \[{{n}^{th}}\] term of the given sequence.

Therefore, \[a{{r}^{n-1}}=128\]

\[\Rightarrow \left( 2 \right){{\left( \sqrt{2} \right)}^{n-1}}=128\]

\[\Rightarrow \left( 2 \right){{\left( 2 \right)}^{\frac{n-1}{2}}}={{\left( 2 \right)}^{7}}\]

\[\Rightarrow {{\left( 2 \right)}^{\frac{n-1}{2}+1}}={{\left( 2 \right)}^{7}}\]

\[\Rightarrow \frac{n-1}{2}+1=7\]

\[\Rightarrow \frac{n-1}{2}=6\]

\[\Rightarrow n-1=12\]

\[\Rightarrow n=13\]

Therefore, \[128\] is the \[{{13}^{th}}\] term of the given sequence.

\[\sqrt{3},3,3\sqrt{3}...\] is \[729\] ?

\[\sqrt{3},3,3\sqrt{3}...\] is the given sequence.

The first term of the G.P. \[a=\sqrt{3}\] and the common ratio \[r={3}/{\sqrt{3}}\;=\sqrt{3}\] .

\[729\] is the \[{{n}^{th}}\] term of the given sequence.

Therefore, \[a{{r}^{n-1}}=729\]

\[\Rightarrow \left( \sqrt{3} \right){{\left( \sqrt{3} \right)}^{n-1}}=729\]

\[\Rightarrow {{\left( 3 \right)}^{{1}/{2}\;}}{{\left( 2 \right)}^{\frac{n-1}{2}}}={{\left( 3 \right)}^{6}}\]

\[\Rightarrow {{\left( 3 \right)}^{\frac{1}{2}+\frac{n-1}{2}}}={{\left( 3 \right)}^{6}}\]

\[\Rightarrow \frac{1}{2}+\frac{n-1}{2}=6\]

\[\Rightarrow \frac{1+n-1}{2}=6\]

\[\Rightarrow \frac{n}{2}=6\]

\[\Rightarrow n=12\]

Therefore, \[729\] is the \[{{12}^{th}}\] term of the given sequence.

\[\frac{1}{3},\frac{1}{9},\frac{1}{27},...\] is \[\frac{1}{19683}\] ?

\[\frac{1}{3},\frac{1}{9},\frac{1}{27},...\] is the given sequence.

The first term of the G.P. \[a=\frac{1}{3}\] and the common ratio \[r=\frac{1}{9}\div \frac{1}{3}=\frac{1}{3}\] .

\[\frac{1}{19683}\] is the \[{{n}^{th}}\] term of the given sequence.

Therefore, \[a{{r}^{n-1}}=\frac{1}{19683}\]

\[\Rightarrow \left( \frac{1}{3} \right){{\left( \frac{1}{3} \right)}^{n-1}}=\frac{1}{19683}\]

\[\Rightarrow {{\left( \frac{1}{3} \right)}^{n}}={{\left( \frac{1}{3} \right)}^{9}}\]

\[\Rightarrow n=9\]

Therefore, \[\frac{1}{19683}\] is the \[{{9}^{th}}\] term of the given sequence.

6. For what values of \[x\] , the numbers \[-\frac{2}{7},x,-\frac{7}{2}\] are in G.P.?

\[-\frac{2}{7},x,-\frac{7}{2}\] are the given numbers and the common ratio \[=\frac{x}{-{2}/{7}\;}=\frac{-7x}{2}\]

We also know that, common ratio \[=\frac{-{7}/{2}\;}{x}=\frac{-7}{2x}\]

Equating both the common ratios we get

\[\frac{-7x}{2}=\frac{-7}{2x}\]

\[\Rightarrow {{x}^{2}}=\frac{-2\times 7}{-2\times 7}=1\]

\[\Rightarrow x=\sqrt{1}\]

\[\Rightarrow x=\pm 1\]

Therefore, the given numbers will be in G.P. for \[x=\pm 1\] .

7. Find the sum up to \[20\] terms in the geometric progression \[0.15,0.015,0.0015...\]

\[0.15,0.015,0.0015...\] is the given G.P.

The first term of the G.P. \[a=0.15\] and the common ratio \[r=\frac{0.015}{0.15}=0.1\] .

The sum of first \[n\] terms of the G.P. is given by the equation \[{{S}_{n}}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}\] .

Therefore, the sum of first \[20\] terms of the given G.P. is 

 \[{{S}_{20}}=\frac{0.15\left[ 1-{{\left( 0.1 \right)}^{20}} \right]}{1-0.1}\]

\[=\frac{0.15}{0.9}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]\]

\[=\frac{15}{90}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]\]

\[=\frac{1}{6}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]\]

Therefore, the sum up to \[20\] terms in the geometric progression \[0.15,0.015,0.0015...\] is \[\frac{1}{6}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]\] .

8. Find the sum of \[n\] terms in the geometric progression \[\sqrt{7},\sqrt{21},3\sqrt{7}...\]

\[\sqrt{7},\sqrt{21},3\sqrt{7}...\] is the given G.P.

The first term of the G.P. \[a=\sqrt{7}\] and the common ratio \[r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}\] .

The sum of first \[n\] terms of the given G.P. is 

 \[{{S}_{n}}=\frac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-\sqrt{3}}\]

\[=\frac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}\]

\[=\frac{\sqrt{7}\left( \sqrt{3}+1 \right)\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-3}\]

\[=\frac{-\sqrt{7}\left( \sqrt{3}+1 \right)\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{2}\]

\[=\frac{-\sqrt{7}\left( 1+\sqrt{3} \right)}{2}\left[ {{\left( 3 \right)}^{\frac{n}{2}}}-1 \right]\]

Therefore, the sum of \[n\] terms of the geometric progression \[\sqrt{7},\sqrt{21},3\sqrt{7}...\] is \[\frac{-\sqrt{7}\left( 1+\sqrt{3} \right)}{2}\left[ {{\left( 3 \right)}^{\frac{n}{2}}}-1 \right]\] .

9. Find the sum of \[n\] terms in the geometric progression \[1,-a,{{a}^{2}},-{{a}^{3}}...\]( if \[a\ne -1\] )

\[1,-a,{{a}^{2}},-{{a}^{3}}...\] is the given G.P.

The first term of the G.P. \[{{a}_{1}}=1\] and the common ratio \[r=-a\] .

The sum of first \[n\] terms of the G.P. is given by the equation \[{{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r}\] .

 \[{{S}_{n}}=\frac{1\left[ 1-{{\left( -a \right)}^{n}} \right]}{1-\left( -a \right)}\]

\[=\frac{\left[ 1-{{\left( -a \right)}^{n}} \right]}{1+a}\]

Therefore, the sum of \[n\] terms of the geometric progression \[1,-a,{{a}^{2}},-{{a}^{3}}...\] is \[\frac{\left[ 1-{{\left( -a \right)}^{n}} \right]}{1+a}\] .

10. Find the sum of \[n\] terms in the geometric progression \[{{x}^{3}},{{x}^{5}},{{x}^{7}}...\]( if \[a\ne -1\] )

\[{{x}^{3}},{{x}^{5}},{{x}^{7}}...\] is the given G.P.

The first term of the G.P. \[a={{x}^{3}}\] and the common ratio \[r={{x}^{2}}\] .

 \[{{S}_{n}}=\frac{{{x}^{3}}\left[ 1-{{\left( {{x}^{2}} \right)}^{n}} \right]}{1-{{x}^{2}}}\]

\[=\frac{{{x}^{3}}\left[ 1-{{x}^{2}}^{n} \right]}{1-{{x}^{2}}}\]

Therefore, the sum of \[n\] terms of the geometric progression \[{{x}^{3}},{{x}^{5}},{{x}^{7}}...\] is \[\frac{{{x}^{3}}\left[ 1-{{x}^{2}}^{n} \right]}{1-{{x}^{2}}}\] .

11. Evaluate \[\sum\limits_{k=1}^{11}{\left( 2+3k \right)}\]

\[\sum\limits_{k=1}^{11}{\left( 2+3k \right)=\sum\limits_{k=1}^{11}{(2)}}+\sum\limits_{k=1}^{11}{(3k)=22+\sum\limits_{k=1}^{11}{\left( {{3}^{k}} \right)}}\] …(1)

We know that,

\[\sum\limits_{k=1}^{11}{({{3}^{k}})={{3}^{1}}+{{3}^{2}}+\ldots +{{3}^{11}}}\]

This sequence \[3,{{3}^{2}},{{3}^{3}},\ldots ,{{3}^{11}}\] forms a G.P. Therefore,

\[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\]

Substituting the values to the above equation we get,

\[\Rightarrow {{S}_{n}}=\frac{3\left[ {{\left( 3 \right)}^{11}}-1 \right]}{\left( 3-1 \right)}\]

\[\Rightarrow {{S}_{n}}=\frac{3}{2}\left( {{3}^{11}}-1 \right)\]

\[\Rightarrow \sum\limits_{k=1}^{11}{{{3}^{k}}}=\frac{3}{2}\left( {{3}^{11}}-1 \right)\]

Substitute this value in equation (1). 

\[\sum\limits_{k=1}^{11}{\left( 2+3k \right)=22+\frac{3}{2}\left( {{3}^{11}}-1 \right)}\]

12. The sum of first three terms of a G.P. is \[\frac{39}{10}\] and their product is \[1\] . Find the common ratio and the terms.

Let the first three terms of a G.P. be \[\frac{a}{r},a,ar\]. 

Then, its sum is

\[\frac{a}{r}+a+ar=\frac{39}{10}\]       …(1)

And the product is

\[\left( \frac{a}{r} \right)\left( a \right)\left( ar \right)=1\]      …(2)

Solving equation (2) we will get, 

\[{{a}^{3}}=1\]

Considering the real roots,

Substitute the value of \[a\] in the equation. 

\[\frac{1}{r}+1+r=\frac{39}{10}\]

\[\Rightarrow 1+r+{{r}^{2}}=\frac{39}{10}r\]

\[\Rightarrow 10+10r+10{{r}^{2}}=39r\]

\[\Rightarrow 10{{r}^{2}}-29r+10=0\]

\[\Rightarrow 10{{r}^{2}}-25r-4r+10=0\]

\[\Rightarrow 5r\left( 2r-5 \right)-2\left( 2r-5 \right)=0\]

\[\Rightarrow \left( 5r-2 \right)\left( 2r-5 \right)=0\]

\[\Rightarrow r=\frac{2}{5}\] or \[\frac{5}{2}\]

Therefore, \[\frac{5}{2},1\] and \[\frac{2}{5}\] are the first three terms of the G.P.

13. How many terms of G.P. \[3,{{3}^{2}},{{3}^{3}}...\] are needed to give the sum 120?

Given G.P. \[3,{{3}^{2}},{{3}^{3}},\ldots ,{{3}^{11}}\] .

Let there be \[n\] terms to get the sum as \[120\]. 

Then using the formula, we get,

\[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\]       …(1)

Given that,

\[{{S}_{n}}=120\]

Substituting the given values in equation (1),

\[{{S}_{n}}=120=\frac{3\left( {{3}^{n}}-1 \right)}{3-1}\]

\[\Rightarrow 120=\frac{3\left( {{3}^{n}}-1 \right)}{2}\]

\[\Rightarrow \frac{120\times 2}{3}={{3}^{n}}-1\]

\[\Rightarrow {{3}^{n}}-1=80\]

\[\Rightarrow {{3}^{n}}=81\]

\[\Rightarrow {{3}^{n}}={{3}^{4}}\]

\[\Rightarrow n=4\]

Therefore, for getting the sum as \[120\] the given G.P. should have \[4\] terms.

14. The sum of first three terms of a G.P. is \[16\] and the sum of the next three terms is \[128\]. Determine the first term, the common ratio, and the sum to \[n\] terms of the G.P.

Let \[a,ar,a{{r}^{2}},a{{r}^{3}}...\] be the G.P.

\[a+ar+a{{r}^{2}}=16\]         …(1)

\[a{{r}^{3}}+a{{r}^{4}}+a{{r}^{5}}=128\]   …(2)

Equation (1) and (2)  can also be written as,

\[a\left( 1+r+{{r}^{2}} \right)=16\]

\[a{{r}^{3}}\left( 1+r+{{r}^{2}} \right)=128\]

Divide equation (2) by (1) .

\[\frac{\left( 2 \right)}{\left( 1 \right)}\Rightarrow \frac{a{{r}^{3}}\left( 1+r+{{r}^{2}} \right)}{a\left( 1+r+{{r}^{2}} \right)}=\frac{128}{16}\]

\[\Rightarrow {{r}^{3}}=8\]

\[\Rightarrow r=2\]

Substituting the value of \[r\] in equation (1), we get

\[\Rightarrow a\left( 1+2+4 \right)=16\]

\[\Rightarrow 7a=16\]

\[\Rightarrow a=\frac{16}{7}\]

Sum of \[n\] terms of the G.P. is,

\[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\]\[\]

\[\Rightarrow {{S}_{n}}=\frac{16}{7}\frac{\left( {{2}^{n}}-1 \right)}{2-1}\]

\[\Rightarrow {{S}_{n}}=\frac{16}{7}\left( {{2}^{n}}-1 \right)\]

Therefore, the first term of the G.P. is \[a=\frac{16}{7}\], the common ratio \[r=2\] and the sum of terms \[{{S}_{n}}=\frac{16}{7}\left( {{2}^{n}}-1 \right)\] .

15. Given a G.P. with \[a=729\] and \[{{7}^{th}}\] term \[64\], determine \[{{S}_{7}}\] .

Given that\[a=729\] and \[{{a}_{7}}=64\]

Let the common ratio of the G.P be \[r\]. Then,

\[{{a}_{n}}=a{{r}^{n-1}}\]

\[\Rightarrow {{a}_{7}}=a{{r}^{6-1}}\]

\[\Rightarrow 64=729\left( {{r}^{6}} \right)\]

\[\Rightarrow {{r}^{6}}={{\left( \frac{2}{3} \right)}^{6}}\]

\[\Rightarrow r=\frac{2}{3}\]

\[{{S}_{n}}=\frac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]

\[{{S}_{7}}=\frac{729\left( 1-{{\left( \frac{2}{3} \right)}^{7}} \right)}{\left( 1-\frac{2}{3} \right)}\]

\[=729\times 3\left( \frac{{{\left( 3 \right)}^{7}}-{{\left( 2 \right)}^{7}}}{{{\left( 3 \right)}^{7}}} \right)\]

\[={{\left( 3 \right)}^{7}}\left( \frac{{{\left( 3 \right)}^{7}}-{{\left( 2 \right)}^{7}}}{{{\left( 3 \right)}^{7}}} \right)\]

\[={{\left( 3 \right)}^{7}}-{{\left( 2 \right)}^{7}}\]

\[=2187-128\]

Therefore, the value of \[{{S}_{7}}\] is \[2059\] . 

16. Find a G.P. for which sum of the first two terms is \[-4\] and the fifth term is \[4\] times the third term.

Let \[a\] and \[r\] be the first term and common ratio of the G.P. respectively.

According to the conditions given in  the question,

\[{{a}_{5}}=4\times {{a}_{3}}\]

\[\Rightarrow a{{r}^{4}}=4\times a{{r}^{2}}\]

\[\Rightarrow {{r}^{2}}=4\]

\[\Rightarrow r=\pm 2\]

\[{{S}_{2}}=-4=\frac{a\left( 1-{{r}^{2}} \right)}{\left( 1-r \right)}\]

Substituting \[r=2\] in the above equation,

\[-4=\frac{a\left[ 1-{{\left( 2 \right)}^{2}} \right]}{1-2}\]

\[\Rightarrow -4=\frac{a\left( 1-4 \right)}{-1}\]

\[\Rightarrow -4=a\left( 3 \right)\]

\[\Rightarrow a=\frac{-4}{3}\]

Now, taking \[r=-2\] , we get,

\[-4=\frac{a\left[ 1-{{\left( -2 \right)}^{2}} \right]}{1-\left( -2 \right)}\]

\[\Rightarrow -4=\frac{a\left( 1-4 \right)}{1+2}\]

\[\Rightarrow -4=\frac{a\left( -3 \right)}{3}\]

\[\Rightarrow a=4\]

Therefore , \[\frac{-4}{3},\frac{-8}{3},\frac{-16}{3},...\] or \[4,-8,-16,-32...\] is the required G.P. 

17. If the \[{{4}^{th}}\],\[{{10}^{th}}\] and \[{{16}^{th}}\]  terms of a G.P. are \[x,y\] and \[z\] , respectively. Prove that \[x,y,z\] are in G.P.

Let the first term of the G.P be \[a\] and the common ratio be \[r\].

\[{{a}_{4}}=a{{r}^{3}}=x\]       …(1)

\[{{a}_{10}}=a{{r}^{9}}=y\]      …(2)

\[{{a}_{16}}=a{{r}^{15}}=z\]      …(3)

Then divide equation (2) by (1) .

\[\frac{y}{x}=\frac{a{{r}^{9}}}{a{{r}^{3}}}\]

\[\Rightarrow \frac{y}{x}={{r}^{6}}\]

Now, divide equation (3) by (1).

\[\frac{z}{y}=\frac{a{{r}^{15}}}{a{{r}^{9}}}\]

\[\Rightarrow \frac{z}{y}={{r}^{6}}\]

Therefore, 

\[\frac{y}{x}=\frac{z}{y}\]

Therefore, it is proved that \[x,y,z\] are in G. P.

18. Find the sum to \[n\] terms of the sequence, \[8,88,888,8888...\]

\[8,88,888,8888...\] is the given sequence  

The given sequence is not in G.P. In order to make the sequence in G.P., it has to be changed to the form,

\[{{S}_{n}}=8+88+888+8888+...\] to \[n\] terms

\[=\frac{8}{9}\](\[9+99+999+9999+...\]to \[n\] terms)

\[=\frac{8}{9}\](\[\left( 10-1 \right)+\left( {{10}^{2}}-1 \right)+\left( {{10}^{3}}-1 \right)+\left( {{10}^{4}}-1 \right)+\]to \[n\] terms)

\[=\frac{8}{9}\](\[10+{{10}^{2}}+...n\] terms)\[-\]( \[1+1+1+...n\] terms)

\[=\frac{8}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{10-1}-n \right]\]

\[=\frac{8}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{9}-n \right]\]

\[=\frac{80}{81}\left( {{10}^{n}}-1 \right)-\frac{8}{9}n\]

Therefore, the sum of \[n\] terms the given sequence is \[\frac{80}{81}\left( {{10}^{n}}-1 \right)-\frac{8}{9}n\] .

19. Find the sum of the products of the corresponding terms of the sequences \[2,4,8,16,32\] and \[128,32,8,2,{1}/{2}\;\] .

\[2\times 128+4\times 32+8\times 8+16\times 2+32\times \frac{1}{2}\] 

\[=64\left[ 4+2+1+\frac{1}{2}+\frac{1}{{{2}^{2}}} \right]\]

is the required sum.

We can see that, \[4,2,1,\frac{1}{2},\frac{1}{{{2}^{2}}}\] is a G.P.

The first term of the G.P. is \[a=4\] and the common ratio is \[r=\frac{1}{2}\] .

\[{{S}_{3}}=\frac{4\left[ 1-{{\left( \frac{1}{2} \right)}^{5}} \right]}{1-\frac{1}{2}}\]

\[=\frac{4\left[ 1-\frac{1}{32} \right]}{\frac{1}{2}}\]

\[=8\left( \frac{32-1}{32} \right)\]

\[=\frac{31}{4}\]

Therefore, the required sum \[=64\left( \frac{31}{4} \right)=\left( 16 \right)\left( 31 \right)=496\] .

20. Show that the products of the corresponding terms of the sequences form \[a,ar,a{{r}^{2}},...a{{r}^{n-1}}\] and \[A,AR,A{{R}^{2}},A{{R}^{n-1}}\]  a G.P. and find the common ratio.

The sequence \[aA,arAR,a{{r}^{2}}A{{R}^{2}},...a{{r}^{n-1}}A{{R}^{n-1}}\]  forms a G.P. is to be proved.

Second term / First term \[=\frac{arAR}{aA}=rR\]

Third term / Second term \[=\frac{a{{r}^{2}}A{{R}^{2}}}{aA}=rR\]

Therefore, the \[aA,arAR,a{{r}^{2}}A{{R}^{2}},...a{{r}^{n-1}}A{{R}^{n-1}}\] forms a G.P. and the common ratio is \[rR\].

21. Find four numbers forming a geometric progression in which third term is greater than the first term by \[9\], and the second term is greater than the \[{{4}^{th}}\] by \[18\] .

Let the first term be \[a\] and the common ratio be \[r\] of the G.P.

\[{{a}_{1}}=a,{{a}_{2}}=ar,{{a}_{3}}=a{{r}^{2}},{{a}_{4}}=a{{r}^{3}}\]

\[{{a}_{3}}={{a}_{1}}+9\]   

\[\Rightarrow a{{r}^{2}}=a+9\]   

\[\Rightarrow a\left( {{r}^{2}}-1 \right)=9\]      ...(1)

 \[{{a}_{2}}={{a}_{4}}+9\]   

\[\Rightarrow ar=a{{r}^{3}}+18\]   

\[\Rightarrow ar\left( 1-{{r}^{2}} \right)=18\]    ...(2)

Divide (2) by (1).

\[\frac{ar\left( 1-{{r}^{2}} \right)}{a\left( {{r}^{2}}-1 \right)}=\frac{18}{9}\]

\[\Rightarrow -r=2\]

\[\Rightarrow r=-2\]

Substitute \[r=-2\] in equation (1).

\[a\left( 4-1 \right)=9\]

\[\Rightarrow a\left( 3 \right)=9\]

\[\Rightarrow a=3\]

Therefore, \[3,3\left( -2 \right),3{{\left( -2 \right)}^{2}}\] and \[3{{\left( -2 \right)}^{3}}\] ,i.e., \[3,-6,12\] and \[-24\] are the first four numbers of the G.P. 

22. If \[{{p}^{th}}\], \[{{q}^{th}}\] and \[{{r}^{th}}\] terms of a G.P. are \[a,b\] and \[c\] , respectively. Prove that \[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}=1\] .

Let the first term be \[A\] and the common ration be \[R\] of the G.P.

\[A{{R}^{p-1}}=a\]

\[A{{R}^{q-1}}=b\]

\[A{{R}^{r-1}}=c\]

\[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}\]

\[={{A}^{q-r}}\times {{R}^{\left( p-1 \right)\left( q-r \right)}}\times {{A}^{r-p}}\times {{R}^{\left( q-1 \right)\left( r-p \right)}}\times {{A}^{q-r}}\times {{R}^{\left( r-1 \right)\left( p-q \right)}}\]

\[={{A}^{q-r+r-p+p-q}}\times {{R}^{\left( pq-pr-q+r \right)+\left( rq-r+p-pq \right)+\left( pr-p-qr+q \right)}}\]

\[={{A}^{0}}\times {{R}^{0}}\]

Therefore, \[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}=1\] is proved.

23. If the first and \[{{n}^{th}}\] the term of a G.P. are \[a\] and \[b\] , respectively, and if \[P\] is the product of \[n\] terms, prove that \[{{P}^{2}}={{\left( ab \right)}^{n}}\] .

\[a\] is the first term and \[b\] is the last term of the G.P.

Therefore, is the G.P. \[a,ar,a{{r}^{2}},a{{r}^{3}}...a{{r}^{n-1}}\] , where the common ratio is \[r\] .

\[b=a{{r}^{n-1}}\]                …(1)

\[P\] is the product of \[n\] terms. Therefore,

\[P=\left( a \right)\left( ar \right)\left( a{{r}^{2}} \right)...\left( a{{r}^{n-1}} \right)\]

\[=\left( a\times a\times ...a \right)\left( r\times {{r}^{2}}\times ...{{r}^{n-1}} \right)\]

\[={{a}^{n}}{{r}^{1+2+...\left( n-1 \right)}}\]          …(2)

We can see that, \[1,2,...\left( n-1 \right)\] is an A.P. Therefore,

\[1+2+...+\left( n-1 \right)\]

\[=\frac{n-1}{2}\left[ 2+\left( n-1-1 \right)\times 1 \right]\]

\[=\frac{n-1}{2}\left[ 2+n-2 \right]\]

\[=\frac{n\left( n-1 \right)}{2}\]

So, equation (2) can be written as \[P={{a}^{n}}{{r}^{\frac{n\left( n-1 \right)}{2}}}\].

\[{{P}^{2}}={{a}^{2n}}{{r}^{n\left( n-1 \right)}}\]

\[={{\left[ {{a}^{2}}{{r}^{\left( n-1 \right)}} \right]}^{n}}\]

\[={{\left[ a\times a{{r}^{n-1}} \right]}^{n}}\]

Substituting (1) in the equation,

\[{{P}^{2}}={{\left( ab \right)}^{n}}\]

Therefore, \[{{P}^{2}}={{\left( ab \right)}^{n}}\] is proved.

24. Show that the ratio of the sum of first \[n\] terms of a G.P. to the sum of terms from \[{{\left( n+1 \right)}^{th}}\] to \[{{\left( 2n \right)}^{th}}\] term is \[\frac{1}{{{r}^{n}}}\] .

Let the first term be \[a\] and the common ration be \[r\] of the G.P.

\[\frac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\] is the sum of first \[n\] terms.

From \[{{\left( n+1 \right)}^{th}}\] to \[{{\left( 2n \right)}^{th}}\] term there are \[n\] terms.

From \[{{\left( n+1 \right)}^{th}}\] to \[{{\left( 2n \right)}^{th}}\] term the sum of the terms is 

\[{{S}_{n}}=\frac{{{a}_{n+1}}\left( 1-{{r}^{n}} \right)}{1-r}\] 

\[{{a}^{n+1}}=a{{r}^{n+1-1}}=a{{r}^{n}}\]

Therefore, the required ratio is \[=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}\times \frac{1-r}{a{{r}^{n}}\left( 1-{{r}^{n}} \right)}=\frac{1}{{{r}^{n}}}\]

Therefore, \[\frac{1}{{{r}^{n}}}\] is the ratio of the sum of first \[n\] terms of a G.P. to the sum of terms from \[{{\left( n+1 \right)}^{th}}\] to \[{{\left( 2n \right)}^{th}}\]  term .

25. If \[a,b,c\] and \[d\] are in G.P. show that: \[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)={{\left( ab+bc+cd \right)}^{2}}\]

Let us assume \[a,b,c,d\] are in G.P.

\[bc=ad\]         …(1)

\[{{b}^{2}}=ac\]           …(2)

\[{{c}^{2}}=bd\]           …(3)

\[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)={{\left( ab+bc+cd \right)}^{2}}\]

\[={{\left( ab+bc+cd \right)}^{2}}\]

Substitute (1) in the equation.

\[={{\left( ab+ad+cd \right)}^{2}}\]

\[={{\left( ab+d\left( a+c \right) \right)}^{2}}\]

\[={{a}^{2}}{{b}^{2}}+2abd\left( a+c \right)+{{d}^{2}}{{\left( a+c \right)}^{2}}\]

\[={{a}^{2}}{{b}^{2}}+2{{a}^{2}}bd+2acbd+{{d}^{2}}\left( {{a}^{2}}+2ac+{{c}^{2}} \right)\]

Substitute (1) and (2) in the equation.

\[={{a}^{2}}{{b}^{2}}+2{{a}^{2}}{{c}^{2}}+2{{b}^{2}}{{c}^{2}}+{{d}^{2}}{{a}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{c}^{2}}\]

\[={{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}+{{d}^{2}}{{a}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{c}^{2}}\]

\[={{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}\times {{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+{{c}^{2}}{{b}^{2}}+{{c}^{2}}\times {{c}^{2}}+{{c}^{2}}{{d}^{2}}\]

Substitute (2) and (3) in the equation and rearrange the terms.

\[={{a}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)+{{b}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)+{{c}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\]

\[=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\]

\[=L.H.S.\]

Therefore, \[L.H.S.=R.H.S.\]

Therefore, \[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)={{\left( ab+bc+cd \right)}^{2}}\] is proved.

26. Insert two numbers between \[3\] and \[81\] so that the resulting sequence is G.P.

Let the two numbers between \[3\] and \[81\] be \[{{G}_{1}}\] and \[{{G}_{2}}\] such that the series, \[3,{{G}_{1}},{{G}_{2}},81\] , forms a G.P.

\[81=\left( 3 \right){{\left( r \right)}^{3}}\]

\[\Rightarrow {{r}^{3}}=27\]

Taking the real roots, we get \[r=3\].

When \[r=3\],

\[{{G}_{1}}=ar=\left( 3 \right)\left( 3 \right)=9\]

\[{{G}_{2}}=a{{r}^{2}}=\left( 3 \right){{\left( 3 \right)}^{2}}=27\]

Therefore, \[9\] and \[27\] are the two required numbers. 

27. Find the value of \[n\] so that \[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}\] may be the geometric mean between \[a\] and \[b\] .

The geometric mean of \[a\] and \[b\] is \[\sqrt{ab}\] .

According to conditions given in the question,

\[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}=\sqrt{ab}\]

Square on both the sides.

\[\frac{{{\left( {{a}^{n+1}}+{{b}^{n+1}} \right)}^{2}}}{{{\left( {{a}^{n}}+{{b}^{n}} \right)}^{2}}}=ab\]

\[\Rightarrow {{a}^{2n+2}}+2{{a}^{n+1}}{{b}^{n+1}}+{{b}^{2n+2}}=\left( ab \right)\left( {{a}^{2n}}+2{{a}^{n}}{{b}^{n}}+{{b}^{2n}} \right)\]

\[\Rightarrow {{a}^{2n+2}}+2{{a}^{n+1}}{{b}^{n+1}}+{{b}^{2n+2}}={{a}^{2n+1}}b+2{{a}^{n+1}}{{b}^{n+1}}+a{{b}^{2n+1}}\]

\[\Rightarrow {{a}^{2n+2}}+{{b}^{2n+2}}={{a}^{2n+1}}b+a{{b}^{2n+1}}\]

\[\Rightarrow {{a}^{2n+2}}-{{a}^{2n+1}}b=a{{b}^{2n+1}}-{{b}^{2n+2}}\]

\[\Rightarrow {{a}^{2n+1}}\left( a-b \right)={{b}^{2n+1}}\left( a-b \right)\]

\[\Rightarrow {{\left( \frac{a}{b} \right)}^{2n+1}}=1={{\left( \frac{a}{b} \right)}^{0}}\]

\[\Rightarrow 2n+1=0\]

\[\Rightarrow n=\frac{-1}{2}\]

28. The sum of two numbers is \[6\] times their geometric mean, show that numbers are in the ratio \[\left( 3+2\sqrt{2} \right):\left( 3-2\sqrt{2} \right)\] .

Let \[a\] and \[b\] be the two numbers. 

\[\sqrt{ab}\] is the geometric mean.

\[a+b=6\sqrt{ab}\]               …(1)

\[\Rightarrow {{\left( a+b \right)}^{2}}=36ab\] 

\[{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab=36ab-4ab=32ab\]

\[\Rightarrow a-b=\sqrt{32}\sqrt{ab}\]

\[=4\sqrt{2}\sqrt{ab}\]                    …(2)

Add (1) and (2).

\[2a=\left( 6+4\sqrt{2} \right)\sqrt{ab}\]

\[\Rightarrow a=\left( 3+2\sqrt{2} \right)\sqrt{ab}\]

Substitute \[a=\left( 3+2\sqrt{2} \right)\sqrt{ab}\] in equation (1) .

\[b=6\sqrt{ab}-\left( 3+2\sqrt{2} \right)\sqrt{ab}\] \[\Rightarrow b=\left( 3-2\sqrt{2} \right)\sqrt{ab}\]

Divide \[a\] by \[b\] .

\[\frac{a}{b}=\frac{\left( 3+2\sqrt{2} \right)\sqrt{ab}}{\left( 3-2\sqrt{2} \right)\sqrt{ab}}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\]

Therefore, it is proved that the numbers are in the ratio \[\left( 3+2\sqrt{2} \right):\left( 3-2\sqrt{2} \right)\] .

29. If \[A\] and \[B\] be A.M. and G.M., respectively between two positive numbers, prove that the numbers are \[A\pm \sqrt{\left( A+G \right)\left( A-G \right)}\] .

Given: The two positive numbers between A.M. and G.M. are \[A\] and \[G\].

Let \[a\] and \[b\] be these two positive numbers.

Therefore, \[AM=A=\frac{a+b}{2}\]      …(1)

\[GM=G=\sqrt{ab}\]                          …(2)

Simplifying (1) and (2) , we get

\[a+b=2A\]         …(3)

\[ab={{G}^{2}}\]              …(4)

Substituting (3) and (4) in the identity,

\[{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab\],

\[{{\left( a-b \right)}^{2}}=4{{A}^{2}}-4{{G}^{2}}=4\left( {{A}^{2}}-{{G}^{2}} \right)\]

\[{{\left( a-b \right)}^{2}}=4\left( A+G \right)\left( A-G \right)\]

\[\left( a-b \right)=2\sqrt{\left( A+G \right)\left( A-G \right)}\]     …(5)

Adding (3) and (5) we get ,

\[2a=2A+2\sqrt{\left( A+G \right)\left( A-G \right)}\]

\[\Rightarrow a=A+\sqrt{\left( A+G \right)\left( A-G \right)}\]

Substitute \[a=A+\sqrt{\left( A+G \right)\left( A-G \right)}\] in equation (3).

\[b=2A-A-\sqrt{\left( A+G \right)\left( A-G \right)}\]

\[=A-\sqrt{\left( A+G \right)\left( A-G \right)}\]

Therefore, \[A\pm \sqrt{\left( A+G \right)\left( A-G \right)}\] are the two numbers.

30. The number of bacteria in a certain culture doubles every hour. If there were \[30\] bacteria present in the culture originally, how many bacteria will be present at the end of \[{{2}^{nd}}\] hour, \[{{4}^{th}}\] hour and \[{{n}^{th}}\] hour?

The number of bacteria after every hour will form a G.P. as it is given that the number of bacteria doubles every hour. 

Given: \[a=30\] and \[r=2\]

\[{{a}_{3}}=a{{r}^{2}}=\left( 30 \right){{\left( 2 \right)}^{2}}=120\]

That is, \[120\] will be the number of bacteria at the end of \[{{2}^{nd}}\] hour.

\[{{a}_{5}}=a{{r}^{4}}=\left( 30 \right){{\left( 2 \right)}^{4}}=480\]

That is, \[480\] will be the number of bacteria at the end of \[{{4}^{th}}\] hour.

\[{{a}_{n+1}}=a{{r}^{n}}=\left( 30 \right){{2}^{n}}\]

Therefore, \[30{{\left( 2 \right)}^{n}}\] will be the number of bacteria at the end of \[{{n}^{th}}\] hour.

31. What will Rs.\[500\] amounts to in \[10\] years after its deposit in a bank which pays annual interest rate of \[10%\] compounded annually?

Rs.\[500\] is the amount deposited in the bank.

The amount \[=\] Rs.\[500\left( 1+\frac{1}{10} \right)=\] Rs.\[500\left( 1.1 \right)\] , at the end of first year.

The amount \[=\] Rs.\[500\left( 1.1 \right)\left( 1.1 \right)\] , at the end of \[{{2}^{nd}}\] year. 

The amount \[=\] Rs.\[500\left( 1.1 \right)\left( 1.1 \right)\left( 1.1 \right)\] , at the end of \[{{3}^{rd}}\] year and so on. 

Therefore, the amount at the end of \[10\] years 

\[=\] Rs.\[500\left( 1.1 \right)\left( 1.1 \right)...\](\[10\]times)

\[=\] Rs.\[500{{\left( 1.1 \right)}^{10}}\]

32. If A.M. and G.M. of roots of a quadratic equation are \[8\] and \[5\] , respectively, then obtain the quadratic equation.

Let \[a\] and \[b\] be the root of the quadratic equation.

\[A.M.=\frac{a+b}{2}=8\]

\[\Rightarrow a+b=16\]          …(1)

\[G.M.=\sqrt{ab}=5\]

\[\Rightarrow ab=25\]             …(2)

The quadratic equation is given by the equation,

\[{{x}^{2}}-x\](Sum of roots) \[+\] (Product of roots) \[=0\]

\[{{x}^{2}}-x\left( a+b \right)+\left( ab \right)=0\]

Substituting (1) and (2) in the equation.

\[{{x}^{2}}-16x+25=0\]

Therefore, \[{{x}^{2}}-16x+25=0\] is the required quadratic equation.

Miscellaneous Exercise

1. If is a function satisfying \[f\left( x+y \right)=f\left( x \right).f\left( y \right)\] for all \[x,y\in N\] , such that \[f\left( 1 \right)=3\] and \[\sum\limits_{x=1}^{n}{f\left( x \right)=120}\] find the value of \[n\].

According to the given conditions in the question,

\[f\left( x+y \right)=f\left( x \right)\times f\left( y \right)\] for all \[x,y,\in N\]

\[f\left( 1 \right)=3\]

Let \[x=y=1\].

\[f\left( 1+1 \right)=f\left( 1+2 \right)=f\left( 1 \right)f\left( 2 \right)=3\times 3=9\]

We can also write 

\[f\left( 1+1+1 \right)=f\left( 3 \right)=f\left( 1+2 \right)=f\left( 1 \right)f\left( 2 \right)=3\times 9=27\]

\[f\left( 4 \right)=f\left( 1+4 \right)=f\left( 1 \right)f\left( 3 \right)=3\times 27=81\]

Both the first term and common ratio of \[f\left( 1 \right),f\left( 2 \right),f\left( 3 \right),...,\]that is \[3,9,27,...,\] that forms s G.P. is equal to \[3\]

We know that, \[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\]

Given that, \[\sum\limits_{k=1}^{n}{f}\left( x \right)=120\] 

\[120=\frac{3\left( {{3}^{n}}-1 \right)}{3-1}\]

\[\Rightarrow 120=\frac{3}{2}\left( {{3}^{n}}-1 \right)\]

\[\Rightarrow {{3}^{n}}-1=80\] 

\[\Rightarrow {{3}^{n}}=80={{3}^{4}}\]

Therefore, \[4\] is the value of \[n\].

2. The sum of some terms of G.P. is \[315\] whose first term and the common ratio are \[5\] and \[2\], respectively. Find the last term and the number of terms .

Let \[315\] be the sum of \[n\] terms of the G.P.

The first term \[a\] of the A.P. is \[5\] and the common difference \[r\] is \[2\].

Substitute the values of \[a\] and \[r\] in the equation

\[315=\frac{5\left( {{2}^{n}}-1 \right)}{2-1}\]

\[\Rightarrow {{2}^{n}}-1=63\]

\[\Rightarrow {{2}^{n}}=63={{\left( 2 \right)}^{2}}\]

\[\Rightarrow n=6\]

Therefore, the \[{{6}^{th}}\] term is the last term of the G.P.

 \[{{6}^{th}}\]term \[=a{{r}^{6-1}}=\left( 5 \right){{\left( 2 \right)}^{5}}=\left( 5 \right)\left( 32 \right)=160\]

Therefore, \[160\] is the last term of the G.P  and the number of terms is \[6\]. 

3. The first term of a G.P. is \[1\] . The sum of the third term and fifth term is \[90\]. Find the common ratio of G.P.

Let the first term of the G.P. be \[a\] and the common ratio be \[r\] .

Then, \[a=1\]

\[{{a}_{3}}=a{{r}^{2}}={{r}^{2}}\]

\[{{a}_{5}}=a{{r}^{4}}={{r}^{4}}\]

\[{{r}^{2}}+{{r}^{4}}=90\]

\[\Rightarrow {{r}^{4}}+{{r}^{2}}-90=0\]

\[\Rightarrow {{r}^{2}}=\frac{-1+\sqrt{1+360}}{2}\]

\[=\frac{-1+\sqrt{361}}{2}\]

\[=-10\] or \[9\]

\[\Rightarrow r=\pm 3\]

Therefore, \[\pm 3\] is the common ratio of the G.P. 

4. The sum of the three numbers in G.P. is \[56\]. If we subtract \[1,7,21\] from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Let \[a,ar\] and \[a{{r}^{2}}\] be the three numbers in G.P.

\[a+ar+a{{r}^{2}}=56\]

\[\Rightarrow a\left( 1+r+{{r}^{2}} \right)=56\]            …(1)

An A.P. is formed by

\[a-1,ar-7,a{{r}^{2}}-21\]

\[\left( ar-7 \right)-\left( a-1 \right)=\left( a{{r}^{2}}-21 \right)-\left( ar-7 \right)b\]

\[\Rightarrow ar-a-6=a{{r}^{2}}-ar-14\]

\[\Rightarrow a{{r}^{2}}-2ar+a=8\]

\[\Rightarrow a{{r}^{2}}-ar-ar+a=8\]

\[\Rightarrow a\left( {{r}^{2}}+1-2r \right)=8\]

\[\Rightarrow a{{\left( {{r}^{2}}-1 \right)}^{2}}=8\]                   …(2)

Equating (1) and (2), we get

\[\Rightarrow 7\left( {{r}^{2}}-2r+1 \right)=1+r+{{r}^{2}}\]

\[\Rightarrow 7{{r}^{2}}-14r+7-1-r-{{r}^{2}}\]

\[\Rightarrow 6{{r}^{2}}-15r+6=0\]

\[\Rightarrow 6{{r}^{2}}-12r-3r+6=0\]

\[\Rightarrow 6\left( r-2 \right)-3\left( r-2 \right)=0\]

\[\Rightarrow \left( 6r-3 \right)\left( r-2 \right)=0\]

Then,\[8,16\] and \[32\] are the three numbers when \[r=2\]  and \[32,16\] and \[8\] are the numbers when \[r=\frac{1}{2}\].  

Therefore, \[8,16\] and \[32\] are the three required numbers in either case. 

5. A G.P. consists of an even number of terms. If the sum of all the terms is \[5\]  times the sum of terms occupying odd places, then find its common ratio.

Ans:  

Let \[{{T}_{1}},{{T}_{2}},{{T}_{3}},{{T}_{4}},...{{T}_{2n}}\] be the G.P.

\[2n\] is the number of terms.

\[{{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+{{T}_{2n}}=5\left[ {{T}_{1}}+{{T}_{3}}+...+{{T}_{2n-1}} \right]\]

\[\Rightarrow {{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+{{T}_{2n}}-5\left[ {{T}_{1}}+{{T}_{3}}+...+{{T}_{2n-1}} \right]=0\]

\[\Rightarrow {{T}_{2}}+{{T}_{4}}+...+{{T}_{2n}}=4\left[ {{T}_{1}}+{{T}_{3}}+...+{{T}_{2n-1}} \right]\]

Let \[a,ar,a{{r}^{2}},a{{r}^{3}}\] be the G.P.

\[\frac{ar\left( {{r}^{n}}-1 \right)}{r-1}=\frac{4\times a\left( {{r}^{n}}-1 \right)}{r-1}\]

\[\Rightarrow ar=4a\]

\[\Rightarrow r=4\]

Therefore, \[4\] is the common ratio of the G.P. 

6: If \[\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}\left( x\ne 0 \right)\] then show that \[a,b,c\] and \[d\] are in G.P.

\[\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}\]

\[\Rightarrow \left( a+bx \right)\left( b-cx \right)=\left( b+cx \right)\left( a-bx \right)\]

\[\Rightarrow ab-acx+{{b}^{2}}x-bc{{x}^{2}}=ab-{{b}^{2}}x+-acx-bc{{x}^{2}}\]

\[\Rightarrow 2{{b}^{2}}x=2acx\]

\[\Rightarrow {{b}^{2}}=ac\]

\[\Rightarrow \frac{b}{a}=\frac{c}{b}\]

It is also given that,

\[\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}\]

\[\Rightarrow \left( b+cx \right)\left( c-dx \right)=\left( b-cx \right)\left( c+dx \right)\]

\[\Rightarrow bc-bdx+{{c}^{2}}x-cd{{x}^{2}}=bc+bdx-{{c}^{2}}x-cd{{x}^{2}}\]

\[\Rightarrow 2{{c}^{2}}x=2bdx\]

\[\Rightarrow {{c}^{2}}=bd\]

\[\Rightarrow \frac{c}{d}=\frac{d}{c}\]

Equating both the results, we get

\[\frac{b}{a}=\frac{c}{b}=\frac{d}{b}\]

Therefore, it is proved that \[a,b,c\] and \[d\] are in G.P.

7. Let \[S\] be the sum, \[P\] the product and \[R\] the sum of reciprocals of terms in a G.P. Prove that \[{{P}^{2}}{{R}^{n}}={{S}^{n}}\] .

Let \[a,ar,a{{r}^{2}},a{{r}^{3}}...a{{r}^{n-1}}\] be the G.P.

\[S=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\] 

\[P={{a}^{n}}\times {{r}^{1+2+...+n-1}}\] 

Since the sum of first \[n\] natural numbers is \[n\frac{\left( n+1 \right)}{2}\]

\[\Rightarrow P={{a}^{n}}{{r}^{\frac{n\left( n-1 \right)}{2}}}\] 

\[R=\frac{1}{a}+\frac{1}{ar}+...+\frac{1}{a{{r}^{n-1}}}\]

\[=\frac{{{r}^{n-1}}+{{r}^{n-2}}+...r+1}{a{{r}^{n-1}}}\]

Since \[1,r,...{{r}^{n-1}}\]forms a G.P.,

\[\Rightarrow R=\frac{1\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\times \frac{1}{a{{r}^{n-1}}}\]            

\[=\frac{{{r}^{n}}-1}{a{{r}^{n-1}}\left( r-1 \right)}\]

\[{{P}^{2}}{{R}^{n}}={{a}^{2n}}{{r}^{n\left( n-1 \right)}}\frac{{{\left( {{r}^{n}}-1 \right)}^{n}}}{{{a}^{n}}{{r}^{n\left( n-1 \right)}}{{\left( r-1 \right)}^{n}}}\]

\[=\frac{{{a}^{n}}{{\left( {{r}^{n}}-1 \right)}^{n}}}{{{\left( r-1 \right)}^{n}}}\]

\[={{\left[ \frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)} \right]}^{n}}\]

\[={{S}^{n}}\]

Therefore, \[{{P}^{2}}{{R}^{n}}={{S}^{n}}\].

8. If \[a,b,c,d\] are in G.P., prove that \[\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\] are in G.P

Given: 

\[a,b,c\] and \[d\] are in G.P.

\[{{b}^{2}}=ac\]

\[{{c}^{2}}=bd\]

To prove: 

\[\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\] are in G.P.

That is, \[{{\left( {{b}^{n}}+{{c}^{n}} \right)}^{2}}=\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\]

Then, 

L.H.S \[={{\left( {{b}^{n}}+{{c}^{n}} \right)}^{2}}\]

\[={{b}^{2n}}+2{{b}^{n}}{{c}^{n}}+{{c}^{2n}}\]

\[={{\left( {{b}^{2}} \right)}^{n}}+2{{b}^{n}}{{c}^{n}}+{{\left( {{c}^{2}} \right)}^{n}}\]

\[={{\left( ac \right)}^{n}}+2{{b}^{n}}{{c}^{n}}+{{\left( bd \right)}^{n}}\]

\[={{a}^{n}}{{c}^{n}}+{{b}^{n}}{{c}^{n}}+{{b}^{n}}{{c}^{n}}+{{b}^{n}}{{d}^{n}}\]

\[={{a}^{n}}{{c}^{n}}+{{b}^{n}}{{c}^{n}}+{{a}^{n}}{{d}^{n}}+{{b}^{n}}{{d}^{n}}\]

\[={{c}^{n}}\left( {{a}^{n}}+{{b}^{n}} \right)+{{d}^{n}}\left( {{a}^{n}}+{{b}^{n}} \right)\]

\[=\left( {{a}^{n}}+{{b}^{n}} \right)\left( {{a}^{n}}+{{d}^{n}} \right)\]

\[{{\left( {{b}^{n}}+{{c}^{n}} \right)}^{2}}=\left( {{a}^{n}}+{{b}^{n}} \right)\left( {{c}^{n}}+{{d}^{n}} \right)\]

Therefore, \[\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right)\] and \[\left( {{c}^{n}}+{{d}^{n}} \right)\] are in G.P.

9. If \[a\] and \[b\] are the roots of \[{{x}^{2}}-3x+p=0\] and \[c,d\] are roots of \[{{x}^{2}}-12x+q=0\], where \[a,b,c,d\] form a G.P. Prove that \[\left( q+p \right):\left( q-p \right)=17:15\] . 

Given: \[a\] and \[b\] are the roots of \[{{x}^{2}}-3x+p=0\].

\[a+b=3\] and \[ab=p\]    …(1)      

We also know that \[c\] and \[d\] are the roots of \[{{x}^{2}}-12x+q=0\].

\[c+d=12\] and \[cd=q\]   …(2)      

Also, \[a,b,c,d\] are in G.P.

Let us take \[a=x,b=xr,c=x{{r}^{2}}\] and \[d=x{{r}^{3}}\].

We get from (1) and (2) that,

\[\Rightarrow x\left( 1+r \right)=3\]

\[x{{r}^{2}}+x{{r}^{3}}=12\]

\[\Rightarrow x{{r}^{2}}+\left( 1+r \right)=12\]

Divide both the equations obtained.

\[\frac{x{{r}^{2}}\left( 1+r \right)}{x\left( 1+r \right)}=\frac{12}{3}\]

\[x=\frac{3}{1+2}=\frac{3}{3}=1\], when \[r=2\] and

\[x=\frac{3}{1-2}=\frac{3}{-1}=-3\], when \[r=-2\].

\[ab={{x}^{2}}r=2\], \[cd={{x}^{2}}{{r}^{5}}=32\] when \[r=2\] and \[x=1\] .

\[\frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}\]

\[\Rightarrow \left( q+p \right):\left( q-p \right)=17:15\]

\[ab={{x}^{2}}r=18\], \[cd={{x}^{2}}{{r}^{5}}=-288\] when \[r=-2\] and \[x=-3\] .

\[\frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}\]

Therefore, it is proved that \[\left( q+p \right):\left( q-p \right)=17:15\]as we obtain the same for both the cases. 

10. The ratio of the A.M and G.M. of two positive numbers \[a\] and \[b\] is \[m:n\]. Show that \[a:b=\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right):\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\] .

Let \[a\] and \[b\] be the two numbers.

The arithmetic mean, A.M \[=\frac{a+b}{2}\] and the geometric mean, G.M \[=\sqrt{ab}\]

\[\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}\]

\[\Rightarrow \frac{{{\left( a+b \right)}^{2}}}{4\left( ab \right)}=\frac{{{m}^{2}}}{{{n}^{2}}}\]

\[\Rightarrow \left( a+b \right)=\frac{4ab{{m}^{2}}}{{{n}^{2}}}\]

\[\Rightarrow \left( a+b \right)=\frac{2\sqrt{ab}m}{n}\]                         …(1)

Using the above equation in the identity \[{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab\] , we obtain

\[{{\left( a-b \right)}^{2}}=\frac{4ab{{m}^{2}}}{{{n}^{2}}}-4ab=\frac{4ab\left( {{m}^{2}}-{{n}^{2}} \right)}{{{n}^{2}}}\]

\[\Rightarrow \left( a-b \right)=\frac{2\sqrt{ab}\sqrt{{{m}^{2}}-{{n}^{2}}}}{n}\]             …(2)

Add equation (1) and (2)

\[2a=\frac{2\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

\[\Rightarrow a=\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

Substitute in (1) the value of \[a\].

\[b=\frac{2\sqrt{ab}}{n}m-\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

\[=\frac{\sqrt{ab}}{n}m-\frac{\sqrt{ab}}{n}\sqrt{{{m}^{2}}-{{n}^{2}}}\]

\[=\frac{\sqrt{ab}}{n}\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

\[a:b=\frac{a}{b}=\frac{\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}{\frac{\sqrt{ab}}{n}\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}=\frac{\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}{\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}\]

Therefore, it is proved that \[a:b=\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right):\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\].

11. Find the sum of the following series up to \[n\] terms:

\[5+55+555+...\]

Let \[{{S}_{n}}=5+55+555...\] to \[n\] terms.

\[=\frac{5}{9}\](9+99+999+ ... to n terms.)

\[=\frac{5}{9}(( 10-1 )+( {{10}^{2}}-1 )+( {{10}^{3}}-1)+...\]to n terms)

\[=\frac{5}{9}((10+{{10}^{2}}+{{10}^{3}}...\]to n terms)-(1+1+ to n terms))

\[=\frac{5}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{10-1}-n \right]\]

\[=\frac{5}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{9}-n \right]\]

\[=\frac{50}{81}\left( {{10}^{n}}-1 \right)-\frac{5n}{9}\]

Therefore, the sum of \[n\] terms of the given series is \[\frac{50}{81}\left( {{10}^{n}}-1 \right)-\frac{5n}{9}\] .

\[.6+.66+.666.+...\]

Let \[{{S}_{n}}=0.6+0.66+0.666+\] to \[n\] terms.

\[=6\] (0.1+0.11+0.111+... to \[n\] terms)

\[=\frac{6}{9}\] (0.9+0.99+0.999+... to \[n\] terms)

\[=\frac{6}{9} (\left( 1-\frac{1}{10} \right)+\left( 1-\frac{1}{{{10}^{2}}} \right)+\left( 1-\frac{1}{{{10}^{3}}} \right))+... \]to n terms

\[=\frac{2}{3}\]((\[1+1+...\] to \[n\] terms)\[-\] \[\frac{1}{10}\] (\[1+\frac{1}{10}+\frac{1}{{{10}^{2}}}\] to \[n\] terms))

\[=\frac{2}{3}( n-\frac{1}{10}\left( \frac{1-{{\left( \frac{1}{10} \right)}^{n}}}{1-\frac{1}{10}} \right) )\]

\[=\frac{2}{3}n-\frac{2}{30}\times \frac{10}{9}\left( 1-{{10}^{-n}} \right)\]

\[=\frac{2}{3}n-\frac{2}{27}\left( 1-{{10}^{-n}} \right)\]

Therefore, the sum of \[n\] terms of the given series is \[\frac{2}{3}n-\frac{2}{27}\left( 1-{{10}^{-n}} \right)\] .

12. Find the \[{{20}^{th}}\] term of the series \[2\times 4+4\times 6+6\times 8+...+n\] terms.

\[2\times 4+4\times 6+6\times 8+...+n\] is the given series,

Therefore the \[{{n}^{th}}\] term \[{{a}_{n}}=2n\times \left( 2n+2 \right)=4{{n}^{2}}+4n\]

\[{{a}_{20}}=4{{\left( 20 \right)}^{2}}+4\left( 20 \right)\]

\[=4\left( 400 \right)+80\]

\[=1600+80\]

Therefore, \[1680\] is the \[{{20}^{th}}\] term of the series. 

13. A farmer buys a used tractor for Rs.\[12000\]. He pays Rs.\[6000\] cash and agrees to pay the balance in annual installments of Rs.\[500\] plus \[12%\] interest on the unpaid amount. How much will be the tractor cost him?

It is given that Rs.\[6000\] is paid in cash by the farmer.

Therefore, the unpaid amount is given by

Rs.\[12000-\] Rs.\[6000=\]Rs.\[6000\]

According to the conditions given in the question, the interest to be paid annually by the farmer is 

\[12%\] of \[6000\] , \[12%\] of \[5500\] , \[12%\] of \[5000...12%\] of \[500\]

Therefore, the total interest to be paid by the farmer

\[=12%\] of \[6000+12%\] of \[5500+12%\] of \[5000+...+12%\] of \[500\]

\[=12%\] of \[\left( 6000+5500+5000+...+500 \right)\]

\[=12%\] of \[\left( 500+1000+1500+...+6000 \right)\]

With both the first term and common difference equal to \[500\], the series \[500,1000,1500...6000\] is an A.P.

Let \[n\] be the number of terms of the A.P. 

\[6000=500+\left( n-1 \right)500\]

\[\Rightarrow 1+\left( n-1 \right)=12\]

Therefore, the sum of the given A.P.

\[=\frac{12}{2}\left[ 2\left( 500 \right)+\left( 12-1 \right)\left( 500 \right) \right]\]

\[=6\left[ 1000+5500 \right]\]

\[=6\left( 6500 \right)\]

\[=12%\] of Rs.\[39000\]

\[=\] Rs.\[4680\] 

Therefore, the total cost of tractor

\[=\](Rs.\[12000+\]Rs.\[4680\])

\[=\]Rs.\[16680\]

Therefore, the total cost of the tractor is Rs.\[16680\].

14. Shamshad Ali buys a scooter for Rs.\[22000\]. He pays Rs.\[4000\] cash and agrees to pay the balance in annual installment of Rs.\[1000\] plus \[10%\] interest on the unpaid amount. How much will the scooter cost him?

It is given that for Rs.\[22000\] Shamshad Ali buys a scooter and Rs.\[4000\] is paid in cash.

Rs.\[22000-\] Rs.\[4000=\]Rs.\[18000\]

According to the conditions given in the question, the interest to be paid annually

\[10%\] of \[18000\] , \[10%\] of \[17000\] , \[10%\] of \[16000...10%\] of \[1000\]

\[=10%\] of \[18000+10%\] of \[17000+10%\] of \[16000+...+10%\] of \[1000\]

\[=10%\] of \[\left( 18000+17000+16000+...+1000 \right)\]

\[=10%\] of \[\left( 1000+2000+3000+...+18000 \right)\]

With both the first term and common difference equal to \[1000\], the series \[1000,2000,3000...18000\] is an A.P.

\[18000=1000+\left( n-1 \right)1000\]

\[\Rightarrow 1+\left( n-1 \right)=18\]

\[\Rightarrow n=18\]

\[=\frac{18}{2}\left[ 2\left( 1000 \right)+\left( 18-1 \right)\left( 1000 \right) \right]\]

\[=9\left[ 2000+17000 \right]\]

\[=9\left( 19000 \right)\]

\[=171000\]

Therefore, the total interest to be paid

\[=10%\] of  Rs.\[171000\]

\[=\] Rs.\[17100\] 

Therefore, the total cost of scooter

\[=\](Rs.\[22000+\]Rs.\[17100\])

\[=\]Rs.\[39100\]

Therefore, the total cost of the scooter is Rs.\[39100\] .

15. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs \[50\] paise to mail one letter. Find the amount spent on the postage when \[{{8}^{th}}\] set of letter is mailed.

\[4,{{4}^{2}},{{...4}^{8}}\] is the number of letters mailed and it forms a G.P.

The first term \[a=4\] , the common ratio \[r=4\] and the number of terms \[n=8\] of the G.P.

We know that the sum of \[n\] terms of a G.P. is 

\[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\]

\[{{S}_{8}}=\frac{4\left( {{4}^{8}}-1 \right)}{4-1}\]

\[=\frac{4\left( 65536-1 \right)}{3}\]

\[=\frac{4\left( 65535 \right)}{3}\]

\[=4\left( 21845 \right)\]

\[50\] paisa is the cost to mail one letter.

Cost of mailing \[87380\] letters \[=\] Rs.\[87380\times \frac{50}{100}\] \[=\] Rs.\[43690\]

Therefore, Rs.\[43690\] is the amount spent when \[{{8}^{th}}\] set of letter is mailed.

16. A man deposited Rs.\[10000\] in a bank at the rate of \[5%\] simple interest annually. Find the amount in \[{{15}^{th}}\] year since he deposited the amount and also calculate the total amount after \[20\] years.

Rs.\[10000\] is deposited by the man in a bank at the rate of \[5%\] simple interest annually

\[=\frac{5}{100}\times \]Rs.\[10000=\]Rs.\[500\]

\[10000+500+500+...+500\] is the interest in \[{{15}^{th}}\] year. (\[500\] is \[14\] added times)

Therefore, the amount in \[{{15}^{th}}\] year 

\[=\]Rs.\[10000+14\times \]Rs.\[500\]

\[=\]Rs.\[10000+\]Rs.\[7000\]

\[=\]Rs.\[17000\]

Rs.\[10000+500+500+...+500\] is the amount after \[20\] years. (\[500\] is \[20\] added times)

Therefore, the amount after \[20\] years

\[=\]Rs.\[10000+20\times \]Rs.\[500\]

\[=\]Rs.\[10000+\]Rs.\[10000\]

\[=\]Rs.\[20000\]

The total amount after \[20\] years is Rs.\[20000\].

17. A manufacturer reckons that the value of a machine, which costs him Rs. \[15625\] , will depreciate each year by \[20%\]. Find the estimated value at the end of \[5\] years.

The cost of the machine is Rs.\[15625\].

Every year machine depreciates by \[20%\].

Therefore, \[80%\] of the original cost ,i.e., \[\frac{4}{5}\] of the original cost is its value after every year.

Therefore, the value at the end of \[5\] years 

\[=15626\times \frac{4}{5}\times \frac{4}{5}\times ...\times \frac{4}{5}\]

\[=5\times 1024\]

Therefore, Rs.\[5120\] is the value of the machine at the end of \[5\] years.

18. \[150\] workers were engaged to finish a job in a certain number of days. \[4\] workers dropped out on second day, \[4\] more workers dropped out on third day and so on. It took \[8\] more days to finish the work. Find the number of days in which the work was completed.

Let the number of days in which \[150\] workers finish the work be \[x\].

\[150x=150+146+142+...\left( x+8 \right)\]terms

With first term \[a=146\], common difference \[d=-4\] and number of turns as \[\left( x+8 \right)\] , the series \[150+146+142+...\left( x+8 \right)\]terms is an A.P. 

\[\Rightarrow 150x=\frac{\left( x+8 \right)}{2}\left[ 2\left( 150 \right)+\left( x+8-1 \right)\left( -4 \right) \right]\]

\[\Rightarrow 150x=\left( x+8 \right)\left[ 150+\left( x+7 \right)\left( -2 \right) \right]\]

\[\Rightarrow 150x=\left( x+8 \right)\left( 150-2x-14 \right)\]

\[\Rightarrow 150x=\left( x+8 \right)\left( 136-2x \right)\]

\[\Rightarrow 75x=\left( x+8 \right)\left( 68-x \right)\]

\[\Rightarrow 75x=68x-{{x}^{2}}+544-8x\]

\[\Rightarrow {{x}^{2}}+75x-60x-544=0\]

\[\Rightarrow {{x}^{2}}+15x-544=0\]

\[\Rightarrow {{x}^{2}}+32x-17x-544=0\]

\[\Rightarrow x\left( x+32 \right)-17\left( x+32 \right)=0\]

\[\Rightarrow \left( x-17 \right)\left( x+32 \right)=0\]

\[\Rightarrow x=17\] or \[x=-32\]

We know that \[x\] cannot be negative.

So, \[x=17\].

Therefore, \[17\] is the number of days in which the work was completed. Then the required number of days \[=\left( 17+8 \right)=25\] .

Also you can Find the Solutions of all the Maths Chapters Below.

8.1 introduction:.

The word “sequence” is used in much the same way as it is in ordinary English. When we say that a collection of objects is listed in a sequence, we usually mean that the collection is ordered in such a way that it has an identified first member, second member, third member and so on.

8.2 Sequences:

A sequence is a function whose domain is the set of natural numbers. In simpler terms, it is an arrangement of numbers in a particular order, following a specific rule or pattern.

Types of Sequences:

There are two types of Sequences

Finite Sequence and 

Infinite Sequence

Finite Sequence : A finite sequence is a sequence that has a definite number of terms. The number of terms in a finite sequence is countable and fixed. Once you reach the last term, the sequence ends.

For example:

Consider the sequence 3,6,9,12,15,18. This sequence has exactly 6 terms, so it is a finite sequence.

Infinite sequence: An infinite sequence is a sequence that continues indefinitely without terminating. It has an infinite number of terms and does not have a last term.

Consider the sequence 1, 2, 3, 4, ..., n. This sequence continues without end, so it is an infinite sequence.

8.3 Series:

A series is defined as the sum of the terms of a sequence. If you have a sequence of numbers, then the corresponding series is the result of adding these numbers together.

8.4 Geometric Progression(G.P.):

Geometric Progression (G.P.) is a sequence of numbers where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio.

General form of G.P. is a, ar, ar^2, ar^3,.....

8.4.1 Arithmetic Mean: The Arithmetic Mean (A.M.) of a set of numbers is the sum of the numbers divided by the count of the numbers. It is commonly known as the average.

We can calculate A.M. by using the formula 

A.M. = (a1 + a2 + a3 + ... + an) / n

8.4.2 Geometric Mean: The Geometric Mean (G.M.) of a set of numbers is the nth root of the product of the numbers, where n is the count of the numbers.

We can calculate G.M. by using the formula

G.M. = (a1 * a2 * a3 * ... * an)^(1/n)

8.5 Relationship between A.M. and G.M.:

Inequality: The geometric mean of a set of positive numbers is always less than or equal to the arithmetic mean of the same set of numbers.

G.M. ≤ A.M.

Equality Condition: The equality holds if and only if all the numbers in the set are equal.

Overview of Deleted Syllabus for CBSE Class 11 Maths Chapter - Sequences and Series

Chapter	Dropped Topics
Sequences and Series	8.4 Arithmetic Progression (A.P.)
	8.7 Sum to n terms of special series
	Examples 21, 22 and 24
	Question numbers  - 1,2,3,4,5,6,12,15,16,20,23,24,25,26 from Miscellaneous Exercise
	In summary points 3 and 4

Class 11 Maths Chapter 8: Exercises Breakdown

Exercises	Number of Questions
Exercise 8.1	14 Questions and Solutions
Exercise 8.2	32 Questions and Solutions
Miscellaneous Exercise	18 Questions and Solutions

NCERT Solutions for Class 11 Maths Chapter 8 Solutions Sequences and Series by Vedantu provides a comprehensive guide to understanding different types of sequences such as Geometric Progression (G.P.), along with their corresponding series. Key concepts include finding and understanding the relationship between Arithmetic Mean (A.M.) and Geometric Mean (G.M.). This chapter is crucial as it lays the foundation for higher mathematical concepts and applications in various fields. According to previous exam papers, around 5–6 questions are typically asked from this chapter, emphasizing the importance of mastering these topics for scoring well in exams. Detailed solutions and step-by-step explanations provided by Vedantu help students grasp these concepts effectively, preparing them thoroughly for their exams.

Other Study Material for CBSE Class 11 Maths Chapter 8

S. No	Important Links for Chapter 8 Sequences and Series
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Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

NCERT Solutions Class 11 Chapter-Wise Maths PDF

FAQs on NCERT Solutions for Class 11 Maths Chapter 8 - Sequences and Series

1. What Does Arithmetic Progression Consist of?

Progression is all about a sequence where the terms maintain a specific pattern. In the case of Arithmetic progression, there are two consecutive terms and their difference is constant.

In order to understand the concept better, students should go through Class 11 Maths 8th Chapter and particularly this section. It is relatively easier if one can explore the underlying areas of it. Even though the textbook contains sufficient information regarding the same, you can also opt for study materials to understand the functionalities better.

2. How Many Sections are There in Class 11 Maths Chapter 8 Solutions?

Chapter 8 of Class 11 Maths mainly consists of six important sections apart from the introductory part where the entire chapter has been introduced. It further explains the concepts regarding Sequence, Series, Arithmetic and Geometric Progression, correlation between them and sum related to n terms of special series.

There are four exercises as well to help you with advanced questions to make your exam preparation better. These sections essentially give away small details that are required to solve relevant problems.

3. How to get Full Marks in Sequence and Series Chapter?

The Sequence and Series chapter demands a lot of practice to get rid of conceptual errors since it contains lots of difficult sections. The underlying concepts can be a little tricky in the beginning; however, with the right kind of guidance and advanced techniques, achieving a good score is not that difficult.

If your goal is to attain full marks in every question related to this chapter, you should consider practising different kinds of tricky questions as well. Class 11 Maths NCERT solutions Sequence and Series contain various fast-solving techniques that keep you aware of the type of questions you will be facing in exams.

4. Why are NCERT Solutions Class 11 Maths Chapter 8 Important?

With the use of real-world examples, NCERT Solutions Class 11 Maths Chapter 8 teaches students how to grasp sequences in the most basic way possible. The content in NCERT books is compiled by specialists and is also approved by the CBSE board, making them extremely important. Furthermore, there is a good chance that questions from these NCERT Solutions will appear in the tests. See all the solutions on the Vedantu official website to solve the questions in Chapter 8.

5. Do I Need to Practice all Questions Provided in NCERT Solutions Class 11 Maths Sequences and Series?

For finding the sum of sequences as well as the nth term, the NCERT Solutions Class 11 Maths Sequences and Series requires quite a few formulas. To memorize them, students should practice all of the solved examples, as well as the practice questions to reinforce their understanding of the various types of progressions and, as a result, boost their computation speed. Practicing all the questions from Vedantu will help you in various ways.

6. How Many Questions are there in NCERT Solutions Class 11 Maths Chapter 8 Sequences and Series?

There are 74 problems in NCERT Solutions Class 11 Maths Chapter 8 Sequences and Series and are divided into three categories: easy, fairly easy, and extended format. They are organized into four exercises, each of which focuses on a different aspect of this chapter. These include a wide range of questions relating to A.P. and G.P., allowing students to go deeper into the subject. Chapter 8 is not tough to practice so you can visit Vedantu and start preparing the material.

7. What are the Important Topics Covered in NCERT Solutions Class 11 Maths Chapter 8?

The notion of sequences, arithmetic progression, geometric progression, Fibonacci series, the sum of specific natural number sequences involving squares and cube roots are all explained in NCERT Solutions Class 11 Maths Chapter 8. The issue of arithmetic and geometric mean has also been presented with a logical explanation. All the important topics are crucial from the exam point of view. These important topics need more focus while preparing for the exam. All the preparation material is available on the Vedantu website and the Vedantu app free of cost.

8. How CBSE Students can utilize NCERT Solutions Class 11 Maths Chapter 8 effectively?

After giving the chapter a thorough read, students should make it a habit to practice the solved problems every day. After then, they should begin answering the exercise questions, one area at a time, to obtain sufficient practice and confidence in all of the ideas. Students should make a note of the formulas involved in the topic of arithmetic and geometric progression, or they can refer to the highlight section at the end of the chapter. Students can effectively use the NCERT Solutions Class 11 Maths Chapter 8 in this manner.

NCERT Solutions for Class 11 Maths

• Math Article
• Sequences And Series Class 11

Sequences and Series Class 11

In  Sequences and Series class 11 chapter 9 deals with the study of sequences which follow a specific pattern called progression. In this chapter, the concepts such as  arithmetic progression (A.P),  geometric mean, arithmetic mean, the relationship between A.M. and G.M., special series in forms of the sum to n terms of consecutive natural numbers, sum to n terms of squares and cubes of natural numbers will also be studied.

Sequence and Series Class 11 Concepts

The topics and subtopics covered in chapter 9 – Sequences and Series class 11 concepts are:

• Introduction
• Arithmetic mean
• The general term of a G.P.
• Sum to n terms of a G.P.
• Geometric Mean (G.M.)
• Relationship Between A.M. and G.M.
• Sum to n Terms of Special Series

Sequences and Series Class 11 Notes

The different numbers occurring in any particular sequence are known as terms. The terms of a sequence are denoted by

a 1 , a 2 , a 3 ,….,a n

If a sequence has a finite number of terms then it is known as a finite sequence. A sequence is termed as infinite if it is not having a definite number of terms. The nth term of an AP is given by

a + (n-1) d.

Between any two numbers ‘a’ and ‘b’, n numbers can be inserted such that the resulting sequence is an Arithmetic Progression. A 1 ,  A 2 , A 3 ,……, A n  be n numbers between a and b such that a, A 1  ,  A 2 , A 3 ,……, A n , b is in A.P.

Here, a is the 1st term and b is (n+2)th  term. Therefore,

b = a + d[(n + 2) – 1] = a + d (n + 1).

Hence, common difference (d) = ( b-a)/(n+1)

Now, A 1 = a+d= a+((b-a)/(n+1))

A 2 = a+2d = a + ((2(b-a)/(n+1))

A n = a+nd= a + ((n(b-a)/(n+1))}

The nth term of a geometric progression is given by a n  = ar n-1

Sum of nth term:

where n = number of terms, a = first term and d = common difference

Video Lessons on Sequence and Series

A.p. & it’s properties.

H.P. And It’s Properties

Solved Examples on Sequence and Series

Q.1: If a n  = 2 n , then find the first five terms of the series.

Solution: Given: a n  = 2 n

On substituting n = 1, 2, 3, 4, 5, we get

a 1 = 2 1 = 2

a 2 = 2 2 = 4

a 3 = 2 3 = 8

a 4 = 2 4 = 16

a 5 = 2 5 = 32

Therefore, the required terms are 2, 4, 8, 16, and 32.

Q.2: Find the sum of odd integers from 1 to 2001.

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term, a = 1

Common difference, d = 2

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n – 2 = 2000

2n = 2000 + 2 = 2002

S n = 1001 x 1001

S n = 1002001

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Test: Sequence and Series- Case Based Type Questions - Commerce MCQ

10 questions mcq test - test: sequence and series- case based type questions, shamshad ali buys a scooter for ₹ 22,000. he pays ₹ 4,000 cash and agrees to pay the balance in annual instalments of ₹1000 plus 10% interest on the unpaid amount. q. interest paid by ali on 1st instalment.

• D. none of these

Down payment = ₹4000.

Balance payment = ₹18000

Now, interest on 1st instalment

Shamshad Ali buys a scooter for ₹ 22,000. He pays ₹ 4,000 cash and agrees to pay the balance in annual instalments of ₹1000 plus 10% interest on the unpaid amount. Q. Interest paid by Ali on the 3rd instalment?

Interest on 2nd instalment

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Shamshad Ali buys a scooter for ₹ 22,000. He pays ₹ 4,000 cash and agrees to pay the balance in annual instalments of ₹1000 plus 10% interest on the unpaid amount. Q. Interest paid by Ali on the 2nd instalment?

Shamshad Ali buys a scooter for ₹ 22,000. He pays ₹ 4,000 cash and agrees to pay the balance in annual instalments of ₹1000 plus 10% interest on the unpaid amount.

Q. Total interest paid by Ali is:

= 1800 + 1700 + 1600 + .... + 18 terms

which is an A.P. with a = 1800, d = 1700 – 1800

Therefore, total interest

= 9(3600 – 1700) = 9 × 1900 = 17100

Q. How much will scooter cost him?

= 22000 + 17100

A man deposited ₹10000 in a bank at the rate of 5% simple interest annually.

Q. What is the amount paid by him after 1st year?

Therefore, amount after 1 yr

= 10000 + 500

Q. What is the amount paid by him after 3rd year?

∴ Amount after 3 yr = 10000 + 1500

Q. What is the interest paid by him after 2nd year?

Q. What is the amount paid by him after 15th year?

It is an A.P., where a = 10000 and d = 500

Amount in 15th yr = T 15

= 10000 + 14 × 500

= 10000 + 7000 = ₹17000

Q. What is the amount paid by him after the 20th year?

= 10000 + 10000 = ₹20000

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Important Questions for Sequence and Series- Case Based Type Questions

Sequence and series- case based type questions mcqs with answers, online tests for sequence and series- case based type questions.

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